- What does changing \(c\) do to the graph? - Does a vertical translation change whether \(x = 0\) can be substituted? - Examine the denominator.

1. The denominator cannot equal zero, so \(x = 0\) is excluded from the domain of \(\frac{1}{x}\). 2. Adding a constant \(c\) translates the graph vertically but does not change the denominator. 3. Therefore, \(f(0) = \frac{1}{0} + c\) is undefined for every real value of \(c\). 4. Since a y-intercept would require \(x = 0\), the graph can never cross the y-axis.

No. For every real value of \(c\), the domain is \(\mathbb{R} \setminus \{0\}\), so the graph has no y-intercept.

- An excluded input makes the denominator equal zero. - Substitute the excluded input into the denominator. - Solve the resulting equation for the parameter.

1. An input is excluded when the denominator equals zero. 2. Substitute \(x = 3\) into the denominator: \(2(3) - b = 0\). 3. Solving \(6 - b = 0\) gives \(b = 6\). 4. Check: when \(b = 6\), the denominator \(2x - 6\) equals zero at \(x = 3\).

\(b = 6\)

- What happens if the denominator of a fraction is zero? - Find the input that would make this denominator zero.

1. The denominator must not be zero. 2. Setting \(x-5=0\) gives \(x=5\), which is excluded because division by zero is undefined.

\(x=5\) is not allowed because it makes the denominator zero.

- Check the denominator of the original expression before canceling anything. - Which factor appears in both numerator and denominator? - Use the simplified rule only to locate the height of the missing point.

1. The original denominator is zero at \(x=2\), so this input is not in the domain. 2. Canceling the common factor gives \(g(x)=x+1\) for \(x\neq2\). 3. The corresponding line would have output \(3\) at \(x=2\), so the graph has a hole at \((2,3)\).

Excluded input: \(x=2\). Missing point: \((2,3)\).

- Look closely at how the curve behaves on both sides of the missing input. - Does the output keep moving farther up or down, or does the graph approach one finite height? - How are the positions of the nearby branches different in the two panels?

1. In panel a), the branches grow without bound in opposite vertical directions near \(x=1\), so the graph has a vertical asymptote there. 2. In panel b), the graph follows a straight line except for the open point at \((1,2)\); that is a hole, not a vertical asymptote.

a) Vertical asymptote at \(x=1\). b) Hole at \((1,2)\).

- Find the denominator for each requested input. - Compare the signs of the two denominators. - What happens to a reciprocal when its denominator gets very close to zero?

1. \(f(1.9)=\frac{1}{-0.1}=-10\), while \(f(2.1)=\frac{1}{0.1}=10\). 2. Immediately to the left of \(2\), the denominator is negative and close to zero, so the outputs fall without bound. 3. Immediately to the right of \(2\), the denominator is positive and close to zero, so the outputs rise without bound.

a) \(f(1.9)=-10\) and \(f(2.1)=10\). b) The left branch falls without bound; the right branch rises without bound.

- Identify the highest power of \(x\) in the numerator and denominator. - Compare just those leading terms for very large inputs. - What constant does their ratio approach?

1. The numerator and denominator both have degree \(2\). 2. For very large positive or negative inputs, the leading terms \(6x^2\) and \(2x^2\) dominate. Their ratio is \(\frac{6}{2}=3\). 3. The horizontal asymptote is \(y=3\).

\(y=3\).

- A fraction can equal zero. Which part of the fraction can make it undefined? - Substitute each proposed input into both the numerator and denominator. - Can a valid rational-function output be zero?

1. A zero numerator is allowed when the denominator is nonzero. Thus, \(f(4)=\frac{0}{3}=0\). 2. At \(x=1\), the denominator equals zero, so the function is undefined there. 3. The second student is correct.

The second student is correct. \(x=1\) is excluded, and \(f(4)=0\).

- Look for the x-value the two branches approach but never reach. - Read that position from the horizontal scale. - What do we call a vertical line that the graph approaches without meeting?

1. Both branches move sharply up or down as their x-values get close to \(-2\), but the graph has no point at \(x=-2\). 2. The missing input is \(x=-2\), where the graph has a vertical asymptote.

\(x=-2\); a vertical asymptote.

- Compare the outputs for the largest-magnitude inputs with a single nearby whole number. - Are the negative-input values a little less or a little greater than that number? - Repeat the comparison for positive inputs.

1. The negative-input outputs get closer to \(4\) from below as the inputs become more negative. 2. The positive-input outputs get closer to \(4\) from above as the inputs become more positive. 3. The table suggests the horizontal asymptote \(y=4\).

\(y=4\). For the listed negative inputs, outputs are below \(4\); for the listed positive inputs, they are above \(4\).

- Substitute \(x = 0\) to describe a \(y\)-intercept. - Solve the resulting equation for the parameter. - A \(y\)-intercept cannot exist when \(x = 0\) is not in the domain. - Determine when the denominator is zero at \(x = 0\).

1. For the graph to cross the \(y\)-axis at \((0, 3)\), require \(f(0) = 3\): \(\frac{12}{k} = 3\). 2. Then \(12 = 3k\), so \(k = 4\). 3. The graph has no \(y\)-intercept when \(x = 0\) is excluded from the domain. 4. The denominator is zero at \(x = 0\) when \(0 + k = 0\), so \(k = 0\). In that case, the function is undefined on the \(y\)-axis.

a) \(k = 4\) b) \(k = 0\), because the denominator is zero at \(x = 0\) and the function is undefined there.

- Which input makes the denominator zero? - Check what happens to the numerator at that input. - Is there a factor that could cancel the excluded input?

1. The denominator is zero at \(x=-6\), so this input is excluded. The domain is \(\mathbb{R}\setminus\{-6\}\). 2. At \(x=-6\), the numerator is \(8(-6)-4=-52\), which is not zero. No factor cancels the denominator's zero, so the vertical asymptote is \(x=-6\).

Domain: \(\mathbb{R}\setminus\{-6\}\); vertical asymptote: \(x=-6\).

- Which input makes the denominator equal to \(0\)? - What value can the fractional term never equal? - What is the \(x\)-coordinate of every point on the y-axis?

1. The denominator cannot equal \(0\), so \(x - 1 \ne 0\) and \(x \ne 1\). The domain is \(\mathbb{R} \setminus \{1\}\). 2. The vertical asymptote is \(x = 1\). 3. As \(|x|\) becomes large, \(\frac{4}{x - 1} \to 0\), so the horizontal asymptote is \(y = 2\). 4. Because the fractional term never equals \(0\), the function never equals \(2\). The range is \(\mathbb{R} \setminus \{2\}\). 5. For the y-intercept, set \(x = 0\): \(f(0) = \frac{4}{-1} + 2 = -2\). The y-intercept is \((0, -2)\).

a) Domain: \(\mathbb{R} \setminus \{1\}\); range: \(\mathbb{R} \setminus \{2\}\) b) Vertical asymptote: \(x = 1\); horizontal asymptote: \(y = 2\) c) \((0, -2)\)

- A rational function is undefined when its denominator equals zero. - Set each denominator equal to zero and solve. - Use the zero-product property when the denominator is factored.

1. For \(f\), solve \(5x + 20 = 0\). This gives \(x = -4\), so the domain is \(\mathbb{R} \setminus \{-4\}\). 2. For \(g\), solve \(0.4x - 2 = 0\). This gives \(x = 5\), so the domain is \(\mathbb{R} \setminus \{5\}\). 3. For \(h\), solve \(7 - x = 0\). This gives \(x = 7\), so the domain is \(\mathbb{R} \setminus \{7\}\). 4. For \(k\), solve \(x(x + 4.5) = 0\). By the zero-product property, \(x = 0\) or \(x = -4.5\). Therefore, the domain is \(\mathbb{R} \setminus \{-4.5, 0\}\).

a) Excluded input: \(x = -4\); domain: \(\mathbb{R} \setminus \{-4\}\) b) Excluded input: \(x = 5\); domain: \(\mathbb{R} \setminus \{5\}\) c) Excluded input: \(x = 7\); domain: \(\mathbb{R} \setminus \{7\}\) d) Excluded inputs: \(x = -4.5\) and \(x = 0\); domain: \(\mathbb{R} \setminus \{-4.5, 0\}\)

- Find the input that makes each denominator zero. - What value does the fractional term approach for very large positive or negative inputs? - Check why the coefficient multiplying the fraction must be nonzero.

1. For \(f\), the denominator is zero when \(x - 2 = 0\), so the vertical asymptote is \(x = 2\). The constant outside the fraction is \(3\), so the horizontal asymptote is \(y = 3\). 2. For \(g\), the denominator is zero when \(x + 1 = 0\), so the vertical asymptote is \(x = -1\). The constant outside the fraction is \(-5\), so the horizontal asymptote is \(y = -5\). 3. For \(y = \frac{a}{x-h}+k\) with \(a\ne0\), the vertical asymptote is \(x=h\) and the horizontal asymptote is \(y=k\). The restriction \(a\ne0\) ensures the vertical asymptote exists.

a) Vertical asymptote: \(x = 2\); horizontal asymptote: \(y = 3\) b) Vertical asymptote: \(x = -1\); horizontal asymptote: \(y = -5\) For \(y = \frac{a}{x - h} + k\) with \(a\ne0\), the asymptotes are \(x=h\) and \(y=k\).

- Check the original fraction at the input the students disagree about. - When is division by a denominator allowed? - Does simplifying an expression change where the original expression was defined?

1. Check where the original expression is defined before canceling factors. 2. The denominator \(x-2\) equals zero at \(x=2\), so that input is excluded. 3. For every \(x\ne2\), the fraction simplifies to \(1\). This does not give a value at \(x=2\). 4. Maya is correct. The domain is \(\mathbb{R}\setminus\{2\}\); the graph is the line \(y=1\) with a hole at \((2,1)\).

Maya is correct. The domain is \(\mathbb{R}\setminus\{2\}\).

- Identify any input that makes the denominator zero. - Evaluate the function carefully at each given input. - Consider what happens to a fraction with a fixed numerator as its denominator grows.

1. Since division by zero is undefined, \(x \ne 0\). Therefore, the domain is \(\mathbb{R} \setminus \{0\}\). 2. \(f(10) = \frac{4}{10} + 3 = 0.4 + 3 = 3.4\). 3. \(f(100) = \frac{4}{100} + 3 = 0.04 + 3 = 3.04\). 4. \(f(1000) = \frac{4}{1000} + 3 = 0.004 + 3 = 3.004\). 5. As \(x\) becomes increasingly large, \(\frac{4}{x}\) approaches \(0\), so \(f(x)\) approaches \(3\) from above. The horizontal asymptote is \(y = 3\).

a) \(\mathbb{R} \setminus \{0\}\) b) \(f(10) = 3.4\), \(f(100) = 3.04\), and \(f(1000) = 3.004\) c) The function values approach \(3\) from above; the horizontal asymptote is \(y = 3\).

- Substitute each given input into the function rule. - What function value defines a zero? - What happens to a fraction with fixed numerator when its denominator grows without bound?

1. Substitute the given inputs: \(f(1) = \frac{6}{1} - 3 = 3\), \(f(2) = \frac{6}{2} - 3 = 0\), \(f(3) = \frac{6}{3} - 3 = -1\), and \(f(6) = \frac{6}{6} - 3 = -2\). 2. Solve \(f(x) = 0\): \(\frac{6}{x} - 3 = 0\), so \(\frac{6}{x} = 3\), \(6 = 3x\), and \(x = 2\). 3. As \(x \to \infty\), \(\frac{6}{x} \to 0\). Therefore, \(f(x) \to -3\).

a) \(f(1) = 3\), \(f(2) = 0\), \(f(3) = -1\), and \(f(6) = -2\) b) \(x = 2\) c) The function values approach \(-3\), because \(\frac{6}{x}\) approaches \(0\).

- Find the input that makes the denominator zero. - Consider what happens to a fraction when the denominator grows without bound. - The intersection of a vertical line \(x = a\) and a horizontal line \(y = b\) is \((a, b)\).

1. The denominator cannot equal zero. Solving \(x + 5 = 0\) gives \(x = -5\). Therefore, the domain is \(\mathbb{R} \setminus \{-5\}\), and the vertical asymptote is \(x = -5\). 2. As \(x \to \pm\infty\), \(\frac{4}{x + 5} \to 0\). Therefore, \(f(x) \to -2\), and the horizontal asymptote is \(y = -2\). 3. The lines \(x = -5\) and \(y = -2\) intersect at \((-5, -2)\).

a) Domain: \(\mathbb{R} \setminus \{-5\}\); vertical asymptote: \(x = -5\) b) Horizontal asymptote: \(y = -2\) c) \((-5, -2)\)

- Which input makes the denominator equal to \(0\)? - What values identify the vertical and horizontal translations of the reciprocal parent function? - What input gives the y-intercept? - What must be true for a point to lie on the graph?

1. The denominator cannot equal \(0\). Since \(x+4=0\) when \(x=-4\), the domain is \(\mathbb{R}\setminus\{-4\}\). 2. The vertical asymptote is \(x=-4\), and the horizontal asymptote is \(y=-3\). 3. Set \(x=0\): \(f(0)=\frac{2}{4}-3=0.5-3=-2.5\). The y-intercept is \((0,-2.5)\). 4. Evaluate at \(x=-2\): \(f(-2)=\frac{2}{2}-3=-2\). Thus \(P(-2,-2)\) lies on the graph.

a) \(\mathbb{R}\setminus\{-4\}\) b) Vertical asymptote: \(x=-4\); horizontal asymptote: \(y=-3\) c) \((0,-2.5)\) d) Yes, because \(f(-2)=-2\).

- What happens when a denominator equals \(0\)? - Can a fraction with a nonzero numerator ever equal \(0\)? - Which lines are the asymptotes of a translated reciprocal graph?

1. The denominator cannot equal \(0\). Since \(x + 2 = 0\) when \(x = -2\), the domain is \(\mathbb{R} \setminus \{-2\}\). 2. The expression \(-\frac{4}{x + 2}\) cannot equal \(0\), so \(f(x)\) cannot equal \(1\). The range is \(\mathbb{R} \setminus \{1\}\). 3. Division by \(0\) is undefined, which explains why \(x = -2\) must be excluded. 4. The vertical asymptote is \(x = -2\), and the horizontal asymptote is \(y = 1\).

a) Domain: \(\mathbb{R} \setminus \{-2\}\); range: \(\mathbb{R} \setminus \{1\}\) b) At \(x = -2\), the denominator is \(0\), and division by \(0\) is undefined. c) Vertical asymptote: \(x = -2\); horizontal asymptote: \(y = 1\)

- Find the input that makes each denominator zero. - The constant outside each fraction gives the horizontal asymptote. - Compare the signs of both coordinates of the two intersection points.

1. For \(f\), the vertical asymptote is \(x = 2\) and the horizontal asymptote is \(y = 3\). 2. For \(g\), the vertical asymptote is \(x = -2\) and the horizontal asymptote is \(y = -3\). 3. The asymptotes of \(f\) intersect at \((2, 3)\), and the asymptotes of \(g\) intersect at \((-2, -3)\). 4. These points are reflections of each other across the origin.

a) For \(f\): \(x = 2\) and \(y = 3\). For \(g\): \(x = -2\) and \(y = -3\). b) The intersection points are \((2, 3)\) and \((-2, -3)\); they are symmetric about the origin.

- Find where the denominator is zero. - The excluded input gives the vertical asymptote. - To find the y-intercept, substitute \(x = 0\).

1. The denominator cannot equal zero. Solving \(x + 3 = 0\) gives \(x = -3\), so the domain is \(\mathbb{R} \setminus \{-3\}\). 2. The vertical asymptote is \(x = -3\). As \(|x|\) becomes large, \(\frac{4}{x + 3}\) approaches \(0\), so the horizontal asymptote is \(y = -2\). 3. For the y-intercept, set \(x = 0\): \(f(0) = \frac{4}{3} - 2 = \frac{4}{3} - \frac{6}{3} = -\frac{2}{3}\). 4. The y-intercept is \(\left(0, -\frac{2}{3}\right)\).

a) \(\mathbb{R} \setminus \{-3\}\) b) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = -2\) c) \(\left(0, -\frac{2}{3}\right)\)

- Can you substitute the x-coordinate of the vertical asymptote into the function? - Set the function equal to the horizontal-asymptote value. - After isolating the fraction, what must hold for the equation to have a solution?

1. The vertical asymptote is \(x=2\). This input is excluded because it makes the denominator zero, so the graph has no point with x-coordinate \(2\). 2. Set \(f(x)=1\): \(\frac{3}{x-2}+1=1\). 3. Subtracting \(1\) gives \(\frac{3}{x-2}=0\). This has no solution because the numerator is the nonzero constant \(3\). 4. Therefore, the graph intersects neither asymptote.

a) The function is undefined at \(x=2\), so the graph cannot intersect the vertical asymptote. b) \(\frac{3}{x-2}=0\) has no solution, so the graph cannot intersect the horizontal asymptote.

- What operation is undefined when a denominator equals \(0\)? - Which value is excluded by the denominator? - What does the rational term approach as \(|x|\) grows? - How do the terms \(x+4\) and \(-2\) translate the parent graph?

1. The denominator cannot equal \(0\). Since \(x+4=0\) at \(x=-4\), the domain is \(\mathbb{R}\setminus\{-4\}\). 2. The vertical asymptote is \(x=-4\). 3. As \(|x|\) becomes large, \(\frac{3}{x+4}\) approaches \(0\), so the horizontal asymptote is \(y=-2\). 4. Replacing \(x\) with \(x+4\) translates the graph \(4\) units left, and subtracting \(2\) translates it \(2\) units down.

Domain: \(\mathbb{R}\setminus\{-4\}\). The asymptotes are \(x=-4\) and \(y=-2\). The graph is translated \(4\) units left and \(2\) units down.

- Find the input that makes the denominator zero. - Determine what remains as the fraction approaches \(0\) for large \(|x|\). - Compare the excluded input with the vertical-asymptote equation.

1. The denominator is zero when \(x + 3 = 0\), so \(x = -3\). Therefore, the vertical asymptote is \(x = -3\). 2. As \(|x|\) becomes large, \(\frac{4}{x + 3}\) approaches \(0\), so the horizontal asymptote is \(y = -5\). 3. The domain is \(\mathbb{R} \setminus \{-3\}\). 4. The excluded input \(x = -3\) is the x-value of the vertical asymptote because the function grows without bound as inputs approach that value.

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = -5\) b) Domain: \(\mathbb{R} \setminus \{-3\}\). The excluded input is the x-value of the vertical asymptote.

- Find the input that would make the denominator zero. - Substitute \(x = 0\) to find the \(y\)-intercept. - Set \(f(x) = 0\) to find the \(x\)-intercept.

1. The denominator cannot equal zero, so \(x - 3 \neq 0\). Therefore, \(x \neq 3\), and the domain is \(D = \mathbb{R} \setminus \{3\}\). 2. To find the \(y\)-intercept, substitute \(x = 0\): \(f(0) = \frac{10}{0 - 3} - 2 = -\frac{10}{3} - \frac{6}{3} = -\frac{16}{3}\). Thus, the \(y\)-intercept is \(\left(0, -\frac{16}{3}\right)\). 3. To find the \(x\)-intercept, set \(f(x) = 0\): \(0 = \frac{10}{x - 3} - 2\). Then \(2 = \frac{10}{x - 3}\), so \(2(x - 3) = 10\). Thus, \(x = 8\), and the \(x\)-intercept is \((8, 0)\).

Domain: \(D = \mathbb{R} \setminus \{3\}\) Intercepts: \(\left(0, -\frac{16}{3}\right)\) and \((8, 0)\)

- Find the input that makes the denominator zero. - At an \(x\)-intercept, the function value is \(0\). - To test a point, substitute its \(x\)-coordinate and compare the output with its \(y\)-coordinate.

1. The denominator cannot equal zero. Solve \(2x - 5 = 0\): \(x = 2.5\). Therefore, the domain is \(D = \mathbb{R} \setminus \{2.5\}\). 2. To find the \(x\)-intercept, set \(f(x) = 0\): \(0 = \frac{3}{2x - 5} + 1\). Then \(\frac{3}{2x - 5} = -1\), so \(3 = -(2x - 5)\). Thus, \(3 = -2x + 5\), giving \(x = 1\). The \(x\)-intercept is \((1, 0)\). 3. Evaluate \(f(4)\): \(f(4) = \frac{3}{2 \cdot 4 - 5} + 1 = \frac{3}{3} + 1 = 2\). 4. Since the calculated value matches the point's \(y\)-coordinate, \(P(4, 2)\) lies on the graph.

a) \(D = \mathbb{R} \setminus \{2.5\}\) b) The \(x\)-intercept is \((1, 0)\). c) Yes. Since \(f(4) = 2\), the point \(P(4, 2)\) lies on the graph.

- A variable denominator may exclude an input; a nonzero constant denominator does not. - A rational expression equals \(0\) when its numerator equals \(0\) and its denominator does not. - Substitute \(x = 0\) to find each \(y\)-intercept.

1. For \(h\), the denominator cannot equal zero. Since \(x + 1 = 0\) at \(x = -1\), the domain is \(D_h = \mathbb{R} \setminus \{-1\}\). 2. The denominator of \(k\) is the nonzero constant \(2\), so \(D_k = \mathbb{R}\). 3. For both functions, the numerator is zero when \(x - 3 = 0\), so \(x = 3\). Since both functions are defined at \(x = 3\), each has the \(x\)-intercept \((3, 0)\). 4. Evaluate each function at \(x = 0\): \(h(0) = \frac{-3}{1} = -3\), while \(k(0) = \frac{-3}{2}\). 5. Therefore, the \(y\)-intercepts are \((0, -3)\) and \(\left(0, -\frac{3}{2}\right)\), which are different.

a) \(D_h = \mathbb{R} \setminus \{-1\}\) and \(D_k = \mathbb{R}\). b) Both functions have the \(x\)-intercept \((3, 0)\). Their \(y\)-intercepts are \((0, -3)\) and \(\left(0, -\frac{3}{2}\right)\).

- Find the input that makes the denominator zero. - Determine the value approached as the fraction tends to \(0\). - A y-intercept has x-coordinate \(0\).

1. The denominator is zero when \(x - 4 = 0\), so the vertical asymptote is \(x = 4\). 2. As \(|x|\) becomes large, \(\frac{2}{x - 4}\) approaches \(0\), so the horizontal asymptote is \(y = 1\). 3. For the y-intercept, set \(x = 0\): \(f(0) = \frac{2}{-4} + 1 = -0.5 + 1 = 0.5\). 4. The y-intercept is \((0, 0.5)\).

a) Vertical asymptote: \(x = 4\); horizontal asymptote: \(y = 1\) b) \((0, 0.5)\)

- Find the input that makes the denominator zero. - The constant outside the fraction determines the horizontal asymptote. - Substitute \(x = 0\) for the \(y\)-intercept. - Set \(f(x) = 0\) for the \(x\)-intercept.

1. The denominator is zero when \(x + 3 = 0\), so the vertical asymptote is \(x = -3\). 2. The constant outside the fraction is \(-2\), so the horizontal asymptote is \(y = -2\). 3. Substitute \(x = 0\): \(f(0) = \frac{6}{3} - 2 = 0\). Therefore, the \(y\)-intercept is \((0, 0)\). 4. Set \(f(x) = 0\): \(0 = \frac{6}{x + 3} - 2\). Then \(2 = \frac{6}{x + 3}\), so \(2(x + 3) = 6\). Thus, \(x = 0\), and the \(x\)-intercept is also \((0, 0)\).

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = -2\). b) The \(y\)-intercept is \((0, 0)\). c) The \(x\)-intercept is \((0, 0)\).

- What happens to the fractional term at very large positive or negative inputs? - Locate the vertical line about which the graph is symmetric. - Replace \(x\) with \(-x\) and compare the two expressions on their common domain.

1. For very large positive or negative inputs, \(\frac{8}{(x+d)^2}\) approaches \(0\), so the horizontal asymptote is \(y=e\). Thus \(e=2\). 2. The graph is symmetric about the vertical line \(x=-d\). For this to be the y-axis, \(-d=0\), so \(d=0\). 3. The resulting function is \(f(x)=\frac{8}{x^2}+2\), with domain \(x\ne0\). The symmetry condition is \(f(-x)=f(x)\). Indeed, \(f(-x)=\frac{8}{(-x)^2}+2=\frac{8}{x^2}+2=f(x)\).

1) \(e=2\) 2) \(d=0\) 3) \(f(-x)=f(x)\), since \(\frac{8}{(-x)^2}+2=\frac{8}{x^2}+2\).

- Which values of \(x\) would make the denominator zero? - How can you find the zeros of a quadratic expression that has no constant term? - When does an excluded domain value produce a vertical asymptote? - Check whether the numerator is also zero at each excluded value.

1. Factor the denominator: \(x^2+2x=x(x+2)\). It is zero at \(x=0\) and \(x=-2\), so the domain is \(D_g=\mathbb{R}\setminus\{-2, 0\}\). 2. Check the numerator at the excluded values. At \(x=0\), \(2x-1=-1\), and at \(x=-2\), \(2x-1=-5\). Neither value is zero. 3. Because neither denominator factor cancels, both excluded values produce vertical asymptotes. Their equations are \(x=-2\) and \(x=0\).

\(D_g=\mathbb{R}\setminus\{-2, 0\}\) Vertical asymptotes: \(x=-2\) and \(x=0\).

- What is the greatest power of \(x\) in the numerator and denominator? - Dividing each term by that power reveals what stays important for large inputs. - What happens to fractions of the form \(\frac{1}{x}\) and \(\frac{1}{x^2}\) as \(x\) grows?

1. For \(x\ne0\), the numerator is \(x^2\left(10-\frac{5}{x}+\frac{1}{x^2}\right)\), and the denominator is \(x^2\left(2+\frac{3}{x^2}\right)\). 2. Cancel the common factor \(x^2\) to obtain \(f(x)=\frac{10-\frac{5}{x}+\frac{1}{x^2}}{2+\frac{3}{x^2}}\) for \(x\ne0\). 3. For large positive inputs, all terms containing \(\frac{1}{x}\) or \(\frac{1}{x^2}\) approach \(0\). Thus the outputs approach \(\frac{10}{2}=5\), giving horizontal asymptote \(y=5\).

a) \(f(x)=\frac{10-\frac{5}{x}+\frac{1}{x^2}}{2+\frac{3}{x^2}}\) for \(x\ne0\) b) The outputs approach \(5\); horizontal asymptote: \(y=5\).

- Compare the highest powers of the numerator and denominator. - Which polynomial grows faster in magnitude for large inputs? - Is the rule for equal-degree polynomials the same as for unequal degrees?

1. The numerator has degree \(1\), while the denominator has degree \(2\). 2. For very large positive or negative inputs, the denominator grows faster in magnitude, so the fraction approaches \(0\). 3. Thus the horizontal asymptote is \(y=0\). Bram is correct that a horizontal asymptote exists but incorrect about its equation.

Bram is partly correct: there is a horizontal asymptote, but it is \(y=0\), not \(y=3\), because the denominator has higher degree.

- Which power of \(x\) is highest in each denominator? - Divide the top and bottom of each fraction by that power and compare what remains for very large inputs. - Think about how \(\frac{1}{x}\) and \(\frac{1}{x^2}\) behave at both ends.

1. For part a, divide the numerator and denominator by \(x\): \(f(x)=\frac{6-\frac{15}{x}}{\frac{3}{x}-2}\). Both ends approach \(\frac{6}{-2}=-3\), so the horizontal asymptote is \(y=-3\). 2. For part b, divide by \(x^2\): \(g(x)=\frac{\frac{5}{x}+\frac{2}{x^2}}{1-\frac{4}{x^2}}\). Both ends approach \(0\), so the horizontal asymptote is \(y=0\). 3. For part c, divide by \(x^2\): \(h(x)=\frac{1.2-\frac{4}{x}}{0.4+\frac{1}{x^2}}\). Both ends approach \(\frac{1.2}{0.4}=3\), so the horizontal asymptote is \(y=3\).

a) Both ends approach \(-3\); horizontal asymptote: \(y=-3\). b) Both ends approach \(0\); horizontal asymptote: \(y=0\). c) Both ends approach \(3\); horizontal asymptote: \(y=3\).

- When is a rational expression equal to \(0\)? - Compare the degrees and leading coefficients to find the horizontal asymptote. - At an intersection with the asymptote, what value must the function equal?

1. A rational function equals \(0\) when its numerator is \(0\) and its denominator is nonzero. Solve \(x^2-6x=x(x-6)=0\), giving \(x=0\) and \(x=6\). 2. The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients: \(y=\frac{1}{2}\). 3. To find the intersection, solve \(\frac{x^2-6x}{2x^2+12}=\frac{1}{2}\). 4. Cross-multiplication gives \(2x^2-12x=2x^2+12\), so \(-12x=12\) and \(x=-1\).

a) Zeros: \(x=0\) and \(x=6\); horizontal asymptote: \(y=\frac{1}{2}\) b) \(x=-1\)

- Compare the numerator's and denominator's highest powers in each fraction. - Divide each numerator and denominator by the denominator's highest power. - Which terms become negligible for very large positive inputs?

1. For \(g\), the numerator and denominator both have degree \(2\). Divide by \(x^2\) to obtain \(\frac{4-\frac{2}{x}+\frac{1}{x^2}}{2+\frac{5}{x^2}}\). The reciprocal terms approach \(0\), leaving \(\frac{4}{2}=2\). 2. For \(h\), the numerator has degree \(1\) but the denominator has degree \(2\). Dividing by \(x^2\) gives \(\frac{\frac{4}{x}-\frac{2}{x^2}}{2+\frac{5}{x^2}}\), whose value approaches \(0\). 3. Thus the horizontal asymptotes are \(y=2\) for \(g\) and \(y=0\) for \(h\).

\(g(x)\) approaches \(2\), with horizontal asymptote \(y=2\). \(h(x)\) approaches \(0\), with horizontal asymptote \(y=0\).

- Identify values that make a denominator equal to \(0\). - Factor before deciding whether a discontinuity is a vertical asymptote. - Examine the function as \(x\to\pm\infty\). - A constant added outside a fraction shifts its horizontal asymptote.

1. For a), the denominator is zero at \(x=4\), so the vertical asymptote is \(x=4\). Since \(\frac{6}{x-4}\to 0\) as \(x\to\pm\infty\), the horizontal asymptote is \(y=2\). 2. For b), factor the denominator: \(x^2+x=x(x+1)\). Neither factor cancels with \(x-1\), so the vertical asymptotes are \(x=0\) and \(x=-1\). Since the denominator has greater degree than the numerator, the horizontal asymptote is \(y=0\).

a) Vertical asymptote: \(x=4\); horizontal asymptote: \(y=2\) b) Vertical asymptotes: \(x=0\) and \(x=-1\); horizontal asymptote: \(y=0\)

- Follow the curve toward the far right and far left of the graph. - Look for the same height that the curve approaches at both ends. - Write the horizontal line using \(y=\text{constant}\).

1. On the far right of the graph, the y-values get closer to \(2\). 2. On the far left, the y-values also get closer to \(2\). 3. Therefore the horizontal asymptote is \(y=2\).

a) The function values approach \(2\) at both ends of the graph. b) \(y=2\)

- A vertical asymptote is the vertical line the graph approaches as the function values increase or decrease without bound. - A horizontal asymptote is the horizontal line the graph approaches as \(x\) becomes very large in magnitude. - The \(x\)-value of a vertical asymptote is excluded from the domain.

1. For \(f\), the graph approaches the vertical line \(x = 3\) and the horizontal line \(y = 2\). Therefore, the asymptotes are \(x = 3\) and \(y = 2\), and the domain is \(D_f = \mathbb{R} \setminus \{3\}\). 2. For \(g\), the graph approaches the vertical line \(x = -2\) and the horizontal line \(y = -1\). Therefore, the asymptotes are \(x = -2\) and \(y = -1\), and the domain is \(D_g = \mathbb{R} \setminus \{-2\}\).

a) \(f\): vertical asymptote \(x = 3\), horizontal asymptote \(y = 2\); \(g\): vertical asymptote \(x = -2\), horizontal asymptote \(y = -1\). b) \(D_f = \mathbb{R} \setminus \{3\}\); \(D_g = \mathbb{R} \setminus \{-2\}\).

- The excluded input occurs where the denominator would equal zero. - The horizontal line approached by the graph gives the vertical translation. - Substitute both coordinates of a point into the function rule. - Consider what happens to a fraction when its denominator grows without bound.

1. The graph approaches the vertical line \(x = 3\). Therefore, the excluded input is \(x = 3\), the vertical asymptote is \(x = 3\), and \(d = 3\). 2. The graph approaches the horizontal line \(y = 2\), so \(e = 2\). Thus, \(f(x) = \frac{1}{x - 3} + 2\). 3. One possible point is \(P(4, 3)\). Substitution gives \(f(4) = \frac{1}{4 - 3} + 2 = 3\), so the point satisfies the equation. 4. As \(x\) becomes very large, \(\frac{1}{x - 3}\) approaches \(0\). Therefore, \(f(x)\) approaches \(2\) from above.

a) Excluded input: \(x = 3\); vertical asymptote: \(x = 3\); \(d = 3\) b) Horizontal asymptote: \(y = 2\); \(e = 2\) c) One possible verification is \(P(4, 3)\): \(f(4) = \frac{1}{4 - 3} + 2 = 3\). d) As \(x \to \infty\), \(f(x) \to 2\) from above.

- The vertical asymptote occurs where the denominator would be zero. - The horizontal asymptote gives the constant added outside the fraction. - At an x-intercept, the output is \(0\).

1. The graph approaches the vertical line \(x = 1\) and the horizontal line \(y = 1\). 2. The denominator is zero when \(x = d\), so the vertical asymptote \(x = 1\) gives \(d = 1\). The horizontal asymptote gives \(e = 1\). Thus, \(f(x) = \frac{2}{x - 1} + 1\). 3. At an x-intercept, \(f(x) = 0\). Solve \(0 = \frac{2}{x - 1} + 1\). 4. Then \(-1 = \frac{2}{x - 1}\), so \(-(x - 1) = 2\) and \(x = -1\). 5. Therefore, the x-intercept is \((-1, 0)\).

a) Vertical asymptote: \(x = 1\); horizontal asymptote: \(y = 1\) b) \(d = 1\) and \(e = 1\) c) \((-1, 0)\)

- At an \(x\)-intercept, the output is \(0\). - Substitute \(x = 0\) to find the \(y\)-intercept. - Consider what happens to a fraction when its denominator grows without bound. - A horizontal asymptote is the horizontal line the graph approaches.

1. Set \(h(x) = 0\): \(0 = \frac{-3}{x + 1} + 2\). Then \(-2 = \frac{-3}{x + 1}\), so \(-2(x + 1) = -3\). Thus, \(2x + 2 = 3\), giving \(x = 0.5\). The \(x\)-intercept is \((0.5, 0)\). 2. Substitute \(x = 0\): \(h(0) = \frac{-3}{1} + 2 = -1\). The \(y\)-intercept is \((0, -1)\), which agrees with the graph. 3. As \(x\) becomes large and positive, \(\frac{-3}{x + 1}\) approaches \(0\). Therefore, \(h(x)\) approaches \(2\), and the horizontal asymptote is \(y = 2\).

a) The \(x\)-intercept is \((0.5, 0)\). b) The \(y\)-intercept is \((0, -1)\). c) The function values approach \(2\), and the horizontal asymptote is \(y = 2\).

- An asymptote is a line that the graph approaches. - At an \(x\)-intercept, the function value is \(0\). - Isolate the fraction before solving the equation.

1. The graph approaches the vertical line \(x = -3\), so the vertical asymptote is \(x = -3\). It approaches the horizontal line \(y = -1\), so the horizontal asymptote is \(y = -1\). 2. To find the \(x\)-intercept, set \(f(x) = 0\): \(\frac{2}{x + 3} - 1 = 0\). 3. Then \(\frac{2}{x + 3} = 1\), so \(2 = x + 3\) and \(x = -1\). 4. Therefore, the \(x\)-intercept is \((-1, 0)\).

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = -1\). b) The \(x\)-intercept is \((-1, 0)\).

- Look at the constant and fractional parts of the rule separately. - What happens to the fractional part when the denominator gets very large? - Does a nearly flat curve have to be close to the x-axis?

1. As \(x\) becomes very large, the denominator \(x+2\) grows, so \(\frac{4}{x+2}\) approaches \(0\). 2. The constant \(0.5\) remains. Therefore, \(g(x)\) approaches \(0.5\), not \(0\). A curve becoming flatter does not mean it approaches the x-axis. 3. The horizontal asymptote is \(y=0.5\).

a) The y-values approach \(0.5\). Marie is mistaken because the fraction approaches \(0\), but the constant \(0.5\) remains. b) \(y=0.5\)

- Check where the curve rises or falls without bound, and which height it approaches far from the center. - What happens to the denominator and numerator at the excluded input? - Compare the leading terms for large positive or negative inputs. - Which input gives the y-intercept?

1. From the graph, the vertical asymptote is \(x=-2\) and the horizontal asymptote is \(y=3\). 2. The denominator is zero at \(x=-2\), but the numerator is \(-12\), so \(x=-2\) is a vertical asymptote. The numerator and denominator have the same degree. Their leading-coefficient ratio is \(\frac{3}{1}=3\), confirming the horizontal asymptote \(y=3\). 3. Since \(g(0)=\frac{-6}{2}=-3\), the y-intercept is \((0,-3)\).

1) Vertical asymptote: \(x=-2\); horizontal asymptote: \(y=3\) 2) At \(x=-2\) the denominator is zero but the numerator is not; the leading-coefficient ratio is \(3\). 3) \((0,-3)\)

- Compare the highest powers of \(x\) in each numerator and denominator. - If the numerator grows more slowly, what happens to the ratio for large inputs? - If the powers match, compare coefficients; if they do not, consider whether the ratio can approach a constant.

1. In \(f\), the denominator has higher degree than the numerator, so the outputs approach \(0\). The horizontal asymptote is \(y=0\). 2. In \(g\), both degrees are \(2\), so the ratio of leading coefficients is \(\frac{5}{2}\). The horizontal asymptote is \(y=\frac{5}{2}\). 3. In \(h\), the numerator has higher degree than the denominator, so the outputs do not approach a constant. There is no horizontal asymptote.

a) \(y=0\). b) \(y=\frac{5}{2}\). c) No horizontal asymptote.

- Which inputs make the original denominator zero? - Compare the factors of the numerator and denominator. What happens to the domain when a common factor is canceled? - Use the simplified expression for the hole's height and inspect the remaining denominator for the asymptote.

1. The original denominator is zero at \(x=-1\) and \(x=3\), so both inputs are excluded. 2. Canceling the common factor \(x+1\) gives \(r(x)=\frac{x-2}{x-3}\) only when \(x\ne-1,3\). Cancellation does not make the original expression defined at \(x=-1\). 3. At \(x=-1\), the simplified expression equals \(\frac{-3}{-4}=\frac{3}{4}\), so the graph has a hole at \((-1,\frac{3}{4})\). The uncanceled denominator vanishes at \(x=3\), giving a vertical asymptote there. Both agree with the graph.

a) \(x=-1\) and \(x=3\). b) The common factor \(x+1\) makes the hole. The original denominator still excludes \(x=-1\). c) Hole: \((-1,\frac{3}{4})\). Vertical asymptote: \(x=3\).

- What is the sign of the numerator near the excluded input? - Can the squared denominator be negative on either side? - Think about dividing a negative number by a very small positive number.

1. Near \(x=-1\), the numerator \(x-2\) is negative. 2. The denominator \((x+1)^2\) is positive on either side of \(-1\), and gets very close to zero. 3. The quotient is a very large negative number from both sides, so both branches fall without bound.

Both branches fall without bound as \(x\) approaches \(-1\) from either side.

- Compare the highest-degree terms to find the horizontal asymptote. - What equation would an intersection with that horizontal line satisfy? - Check whether the graph lies on different sides of the line around the intersection.

1. The numerator and denominator both have degree \(2\), so the horizontal asymptote is \(y=\frac{2}{1}=2\). 2. Set \(f(x)=2\): \(2x^2+3x+5=2(x^2+1)\), which gives \(3x=-3\), or \(x=-1\). 3. Since \(f(-1)=2\) and the sign of \(f(x)-2=\frac{3(x+1)}{x^2+1}\) changes at \(-1\), the graph crosses at \((-1,2)\).

a) \(y=2\). b) \((-1,2)\). This shows the claim is false.

- What input must the original denominator exclude? - How could the numerator and denominator have a common factor that cancels? - After cancellation, what expression should be left for the allowed inputs?

1. The input \(x=3\) must be excluded, so use a denominator containing \(x-3\). 2. Multiply the line expression by the same factor in the numerator: \(f(x)=\frac{(x-3)(2x-1)}{x-3}\). 3. This simplifies to \(2x-1\) only when \(x\neq3\). At \(x=3\) the line's height would be \(5\), so the missing point is \((3,5)\).

One example is \(f(x)=\frac{(x-3)(2x-1)}{x-3}\), with domain \(\mathbb{R}\setminus\{3\}\).

- Which input makes the denominator equal to \(0\)? - What value can the fractional term never equal? - How do the excluded domain and range values relate to the asymptotes? - For a horizontal translation left, what change is made inside the function input?

1. The denominator cannot equal \(0\), so \(x - 3 \ne 0\) and \(x \ne 3\). The domain is \(\mathbb{R} \setminus \{3\}\). 2. The term \(\frac{4}{x - 3}\) can never equal \(0\), so \(f(x)\) can never equal \(2\). The range is \(\mathbb{R} \setminus \{2\}\). 3. The vertical asymptote is \(x = 3\), and the horizontal asymptote is \(y = 2\). 4. Translating left \(5\) units replaces \(x\) with \(x + 5\), and translating down \(3\) units subtracts \(3\): \(g(x) = \frac{4}{(x + 5) - 3} + 2 - 3 = \frac{4}{x + 2} - 1\).

a) Domain: \(\mathbb{R} \setminus \{3\}\); range: \(\mathbb{R} \setminus \{2\}\) b) Vertical asymptote: \(x = 3\); horizontal asymptote: \(y = 2\) c) \(g(x) = \frac{4}{x + 2} - 1\)

- How do the asymptotes determine \(x_0\) and \(y_0\) in the given form? - How can the known point be used to find \(k\)? - Under what condition can a fraction equal \(0\)? Is that condition possible here?

1. The vertical asymptote \(x = 4\) gives \(x_0 = 4\), and the horizontal asymptote \(y = -1\) gives \(y_0 = -1\). Thus, \(g(x) = \frac{k}{x - 4} - 1\). 2. Substitute \(P(6, 1)\): \(1 = \frac{k}{6 - 4} - 1\). Then \(2 = \frac{k}{2}\), so \(k = 4\). 3. Therefore, \(g(x) = \frac{4}{x - 4} - 1\). 4. If \(g(x) = -1\), then \(-1 = \frac{4}{x - 4} - 1\), which simplifies to \(0 = \frac{4}{x - 4}\). A rational expression with nonzero numerator \(4\) cannot equal \(0\), so there is no solution.

a) \(g(x) = \frac{4}{x - 4} - 1\) b) The value \(-1\) is never attained because \(\frac{4}{x - 4}\) cannot equal \(0\) for any real \(x\).

- Replace \(g(x)\) with \(y\) and solve for \(x\). - Identify any value of \(y\) that makes the resulting expression undefined. - Check directly whether that output value can satisfy the original equation. - Relate the missing output value to the horizontal asymptote.

1. Let \(y = \frac{1 - 3x}{x + 2}\) and solve for \(x\). 2. Multiply by \(x + 2\): \(y(x + 2) = 1 - 3x\). 3. Expand and collect the x-terms: \(yx + 2y = 1 - 3x\), so \(x(y + 3) = 1 - 2y\). 4. For \(y \ne -3\), \(x = \frac{1 - 2y}{y + 3}\). This expression gives a real input for every real \(y\) except \(-3\). 5. When \(y = -3\), the equation becomes \(-3x - 6 = 1 - 3x\), or \(-6 = 1\), which is impossible. Therefore, the range is \(\mathbb{R} \setminus \{-3\}\). 6. The graph approaches the excluded output value, so its horizontal asymptote is \(y = -3\).

Range: \(\mathbb{R} \setminus \{-3\}\); horizontal asymptote: \(y = -3\)

- For equal-degree numerator and denominator, compare the leading coefficients. - A rational expression equals \(0\) when its numerator equals \(0\) and its denominator does not. - Write the intercept as an ordered pair.

1. The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of their leading coefficients. Therefore, \(y = a\). 2. Since the horizontal asymptote is \(y = 5\), it follows that \(a = 5\). 3. The function is \(h(x) = \frac{5x - 15}{x + 3}\). A rational expression equals zero when its numerator equals zero and its denominator does not. 4. Solve \(5x - 15 = 0\): \(x = 3\). The denominator at \(x = 3\) is \(6\), so the function is defined there. 5. Therefore, the \(x\)-intercept is \((3, 0)\).

\(a = 5\), and the \(x\)-intercept is \((3, 0)\).

- Which parameters are determined directly by the asymptotes? - How can the given point be used to find the remaining parameter? - What function value corresponds to an x-intercept?

1. The asymptotes give \(x_0 = 2\) and \(y_0 = -3\), so \(h(x) = \frac{a}{x - 2} - 3\). 2. Substitute \(P(4, -4)\): \(-4 = \frac{a}{4 - 2} - 3\). Then \(-1 = \frac{a}{2}\), so \(a = -2\). Thus, \(h(x) = -\frac{2}{x - 2} - 3\). 3. For an x-intercept, solve \(0 = -\frac{2}{x - 2} - 3\). Then \(3 = -\frac{2}{x - 2}\), so \(3(x - 2) = -2\). This gives \(3x = 4\) and \(x = \frac{4}{3}\). 4. The x-intercept is \(\left(\frac{4}{3}, 0\right)\).

a) \(h(x) = -\frac{2}{x - 2} - 3\) b) Yes. The x-intercept is \(\left(\frac{4}{3}, 0\right)\).

- An excluded input makes the denominator equal zero. - Substitute the given excluded input into the denominator and solve for the parameter. - For part c), substitute \(x = 0\) and check whether the parameter remains in the expression.

1. An input is excluded when the denominator equals zero. Substitute \(x = 3\): \(3a - 12 = 0\). 2. Solving gives \(3a = 12\), so \(a = 4\). 3. With \(a = 4\), the denominator is \(4x - 12\), which is zero only when \(x = 3\). Therefore, the domain is \(\mathbb{R} \setminus \{3\}\). 4. At \(x = 0\), the denominator is \(a(0) - 12 = -12\) for every value of \(a\). Since \(-12 \ne 0\), \(x = 0\) can never be excluded.

a) \(a = 4\) b) \(\mathbb{R} \setminus \{3\}\) c) Substituting \(x = 0\) gives the denominator \(-12\), regardless of \(a\), so it can never equal zero.

- The whole expression is defined only when every fraction in it is defined. - Check each denominator before combining the fractions. - To create one denominator with two excluded inputs, use factors that become zero at those inputs.

1. The term \(\frac{1}{x}\) requires \(x \ne 0\). The term \(\frac{6}{x - 2}\) requires \(x \ne 2\). Therefore, Maya is correct. 2. The domain is \(\mathbb{R} \setminus \{0, 2\}\). 3. A single denominator must contain factors that become zero at both excluded inputs. One possible function is \(g(x) = \frac{1}{x(x - 2)}\). 4. Combining the original fractions also gives \(T(x) = \frac{7x - 2}{x(x - 2)}\), with the same exclusions \(x = 0\) and \(x = 2\).

a) Maya is correct. The excluded inputs are \(x = 0\) and \(x = 2\). b) \(\mathbb{R} \setminus \{0, 2\}\) c) One possible answer is \(g(x) = \frac{1}{x(x - 2)}\).

- Find all four asymptotes first. - Use absolute value to find the distance between parallel lines. - Compare the x-values and y-values of the asymptotes to determine the translation.

1. For \(f\), the vertical asymptote is \(x = 3\) and the horizontal asymptote is \(y = 2\). 2. For \(g\), the vertical asymptote is \(x = -2\) and the horizontal asymptote is \(y = -3\). 3. The distance between the vertical asymptotes is \(|3 - (-2)| = 5\) units. 4. The distance between the horizontal asymptotes is \(|2 - (-3)| = 5\) units. 5. Moving from the asymptotes of \(f\) to those of \(g\) is a translation \(5\) units left and \(5\) units down.

a) For \(f\): \(x = 3\) and \(y = 2\). For \(g\): \(x = -2\) and \(y = -3\). b) Vertical-asymptote distance: \(5\) units; horizontal-asymptote distance: \(5\) units c) Translate \(5\) units left and \(5\) units down.

- Find the input that makes the denominator zero. - Distribute the factor \(2\) to the entire expression. - Compare the denominator and the constant term before and after the multiplication. - Check separately whether each asymptote changes.

1. The denominator of \(f\) is zero when \(2x - 6 = 0\), so \(x = 3\). Therefore, the domain is \(\mathbb{R} \setminus \{3\}\), and the vertical asymptote is \(x = 3\). 2. As \(|x|\) becomes large, \(\frac{3}{2x - 6}\) approaches \(0\), so the horizontal asymptote of \(f\) is \(y = 1\). 3. Multiply the entire function by \(2\): \(g(x) = 2\left(\frac{3}{2x - 6} + 1\right) = \frac{6}{2x - 6} + 2\). 4. The denominator is unchanged, so the vertical asymptote of \(g\) is still \(x = 3\). The horizontal asymptote is \(y = 2\). 5. The statement is false. The horizontal-asymptote value doubles from \(1\) to \(2\), but the vertical asymptote remains \(x = 3\).

a) Domain: \(\mathbb{R} \setminus \{3\}\); asymptotes: \(x = 3\) and \(y = 1\) b) The asymptotes of \(g\) are \(x = 3\) and \(y = 2\). The statement is false because only the horizontal-asymptote value doubles.

- What does the constant outside the fraction tell you about the horizontal asymptote? - How can you use the given point to find the remaining coefficient? - To check a possible function value, substitute that output into the equation and see what would have to be true.

1. The horizontal asymptote of \(h(x) = \frac{a}{x} + c\) is \(y = c\), so \(c = -5\). 2. Substitute \(P(2, -4)\): \(-4 = \frac{a}{2} - 5\). Then \(1 = \frac{a}{2}\), so \(a = 2\). Therefore, \(h(x) = \frac{2}{x} - 5\). 3. Solve \(-4.9 = \frac{2}{x} - 5\). Then \(0.1 = \frac{2}{x}\), so \(x = 20\). 4. To test whether \(h(x) = -5\), solve \(-5 = \frac{2}{x} - 5\). This simplifies to \(0 = \frac{2}{x}\), which has no solution because the numerator is nonzero. 5. Therefore, the function approaches \(-5\) but never equals it.

a) \(a = 2\), \(c = -5\), and \(h(x) = \frac{2}{x} - 5\). b) \(x = 20\) c) No. The equation \(\frac{2}{x} = 0\) has no solution, so the function never equals \(-5\).

- A vertical line through a point uses its x-coordinate. - A horizontal line through a point uses its y-coordinate. - In the translated reciprocal form, the asymptotes determine the translation parameters. - Can a fraction with numerator \(1\) ever equal \(0\)?

1. A vertical line through \((-3, 0)\) has equation \(x = -3\). 2. A horizontal line through \((0, 4.5)\) has equation \(y = 4.5\). 3. In \(h(x) = \frac{1}{x - h_0} + k_0\), the vertical asymptote is \(x = h_0\) and the horizontal asymptote is \(y = k_0\). Therefore, \(h_0 = -3\) and \(k_0 = 4.5\). 4. The function is \(h(x) = \frac{1}{x + 3} + 4.5\). The fraction \(\frac{1}{x + 3}\) can never equal \(0\), so \(h(x)\) can never equal \(4.5\).

a) Vertical asymptote: \(x = -3\); horizontal asymptote: \(y = 4.5\) b) \(h_0 = -3\) and \(k_0 = 4.5\) c) Since \(\frac{1}{x + 3}\) is never \(0\), the function never equals \(4.5\).

- Find where the denominator is zero. - Identify the output value approached for large \(|x|\). - Test values such as \(-1.9\) and \(-1.99\) to determine the one-sided behavior.

1. The denominator is zero when \(x + 2 = 0\), so the vertical asymptote is \(x = -2\). 2. As \(|x|\) becomes large, \(\frac{4}{x + 2}\) approaches \(0\), so the horizontal asymptote is \(y = -3\). 3. Therefore, the domain is \(\mathbb{R} \setminus \{-2\}\). Since \(\frac{4}{x + 2}\) can never equal \(0\), the function can never equal \(-3\), so the range is \(\mathbb{R} \setminus \{-3\}\). 4. As \(x \to -2^+\), the denominator \(x + 2\) is positive and approaches \(0\). Therefore, \(\frac{4}{x + 2}\) grows without bound, so \(f(x) \to +\infty\).

a) Vertical asymptote: \(x = -2\); horizontal asymptote: \(y = -3\) b) Domain: \(\mathbb{R} \setminus \{-2\}\); range: \(\mathbb{R} \setminus \{-3\}\) c) As \(x \to -2^+\), \(f(x) \to +\infty\).

- How do the asymptotes determine \(b\) and \(c\)? - How can the given point be used to find \(a\)? - What equation defines a zero of a function?

1. The vertical asymptote gives \(b = 2\), and the horizontal asymptote gives \(c = -1\). Thus, \(g(x) = \frac{a}{x - 2} - 1\). 2. Substitute \(Q(3, 1)\): \(1 = \frac{a}{3 - 2} - 1\). Therefore, \(1 = a - 1\), so \(a = 2\). 3. The function rule is \(g(x) = \frac{2}{x - 2} - 1\). 4. To find the zero, solve \(0 = \frac{2}{x - 2} - 1\). Then \(1 = \frac{2}{x - 2}\), so \(x - 2 = 2\) and \(x = 4\).

a) \(g(x) = \frac{2}{x - 2} - 1\) b) \(x = 4\)

- Which operation is undefined when a denominator is \(0\)? - For part c), set the function equal to the horizontal-asymptote value. - Can a fraction with a fixed nonzero numerator equal \(0\)?

1. The denominator is \(0\) at \(x = 4\), so the domain is \(\mathbb{R} \setminus \{4\}\). 2. The vertical asymptote is \(x = 4\), and the horizontal asymptote is \(y = 1\). 3. Evaluate \(h(2) = -\frac{2}{2 - 4} + 1 = 2\). 4. An intersection with \(y = 1\) would require \(-\frac{2}{x - 4} + 1 = 1\), or \(-\frac{2}{x - 4} = 0\). A rational expression with nonzero numerator \(-2\) cannot equal \(0\), so the graph never reaches \(y = 1\).

a) Domain: \(\mathbb{R} \setminus \{4\}\); vertical asymptote: \(x = 4\); horizontal asymptote: \(y = 1\) b) \(h(2) = 2\) c) The term \(-\frac{2}{x - 4}\) cannot equal \(0\), so \(h(x)\) cannot equal \(1\).

- How do the asymptotes determine \(x_0\) and \(y_0\)? - Substitute the known point to find the remaining parameter. - When solving for \(x\), first move the term outside the fraction to the other side.

1. The asymptotes give \(x_0 = -2\) and \(y_0 = -3\), so \(k(x) = \frac{a}{x + 2} - 3\). 2. Substitute \(P(0, -2.5)\): \(-2.5 = \frac{a}{2} - 3\). Then \(0.5 = \frac{a}{2}\), so \(a = 1\). 3. The function rule is \(k(x) = \frac{1}{x + 2} - 3\). 4. Solve \(-2 = \frac{1}{x + 2} - 3\). Then \(1 = \frac{1}{x + 2}\), so \(x + 2 = 1\) and \(x = -1\).

The function rule is \(k(x) = \frac{1}{x + 2} - 3\). The function equals \(-2\) when \(x = -1\).

- Rewrite the numerator as a multiple of the denominator plus a remainder. - Read the asymptotes from the translated reciprocal form. - A horizontal translation moves a vertical asymptote by the same amount. - A vertical translation moves a horizontal asymptote by the same amount.

1. Rewrite the numerator as \(2x + 5 = 2(x + 1) + 3\). 2. Then \(f(x) = \frac{2(x + 1) + 3}{x + 1} = 2 + \frac{3}{x + 1}\). 3. Therefore, the vertical asymptote is \(x = -1\), and the horizontal asymptote is \(y = 2\). 4. Translating \(3\) units right moves the vertical asymptote from \(x = -1\) to \(x = 2\). 5. Translating \(2\) units up moves the horizontal asymptote from \(y = 2\) to \(y = 4\).

a) \(f(x) = \frac{3}{x + 1} + 2\); asymptotes: \(x = -1\) and \(y = 2\) b) Translated asymptotes: \(x = 2\) and \(y = 4\)

- Determine which input makes the denominator zero. - Every point on the y-axis has \(x=0\). - Relate the excluded input to the location of the vertical asymptote.

1. The denominator cannot equal zero, so \(x-b\ne0\), which means \(x\ne b\). Therefore, the domain is \(D=\mathbb{R}\setminus\{b\}\). 2. A y-intercept can exist only if \(x=0\) is in the domain. 3. If \(b=0\), then \(0\) is excluded from the domain, so the function has no value at \(x=0\) and the graph cannot cross the y-axis. 4. If \(b\ne0\), then \(0\) is in the domain, and \(h(0)=\frac{1}{-b}+c\) is defined. Therefore, the graph does have a y-intercept. 5. The claim is true. Geometrically, \(x=b\) is the vertical asymptote, so when \(b=0\), the y-axis itself is the vertical asymptote.

The claim is true. The domain is \(D=\mathbb{R}\setminus\{b\}\), so \(x=0\) is excluded exactly when \(b=0\). In that case, the y-axis is the vertical asymptote and the graph has no y-intercept.

- How do the asymptotes determine \(x_0\) and \(y_0\)? - How can the given point be used to find \(k\)? - What input gives the y-intercept? - What function value gives the x-intercept?

1. The asymptotes give \(x_0 = -2\) and \(y_0 = 3\), so \(f(x) = \frac{k}{x + 2} + 3\). 2. Substitute \(P(1, 4)\): \(4 = \frac{k}{1 + 2} + 3\). Then \(1 = \frac{k}{3}\), so \(k = 3\). 3. The function rule is \(f(x) = \frac{3}{x + 2} + 3\). 4. For the y-intercept, set \(x = 0\): \(f(0) = \frac{3}{2} + 3 = 4.5\). The y-intercept is \((0, 4.5)\). 5. For the x-intercept, solve \(0 = \frac{3}{x + 2} + 3\). Then \(-3 = \frac{3}{x + 2}\), so \(x + 2 = -1\) and \(x = -3\). The x-intercept is \((-3, 0)\).

a) \(f(x) = \frac{3}{x + 2} + 3\) b) y-intercept: \((0, 4.5)\); x-intercept: \((-3, 0)\)

- Which parts of the function rule determine the asymptotes? - How can the known point be used to find the remaining parameter? - What input gives the y-intercept?

1. The asymptotes give \(x_0 = -1\) and \(y_0 = 3\), so \(k(x) = \frac{a}{x + 1} + 3\). 2. Substitute \(P(1, 2)\): \(2 = \frac{a}{1 + 1} + 3\). Then \(-1 = \frac{a}{2}\), so \(a = -2\). 3. The function rule is \(k(x) = -\frac{2}{x + 1} + 3\). 4. For the y-intercept, set \(x = 0\): \(k(0) = -\frac{2}{1} + 3 = 1\). The y-intercept is \((0, 1)\).

The function rule is \(k(x) = -\frac{2}{x + 1} + 3\), and the y-intercept is \((0, 1)\).

- Which denominator zero gives the vertical asymptote? - For equal degrees, compare the leading coefficients. - Substitute \(x=0\) to find the y-intercept. - A rational function equals zero when its numerator is zero and its denominator is nonzero.

1. The denominator is zero at \(2x+4=0\), so \(x=-2\). Since the numerator is nonzero there, the vertical asymptote is \(x=-2\). 2. The numerator and denominator have the same degree. The ratio of leading coefficients is \(\frac{3}{2}\), so the horizontal asymptote is \(y=\frac{3}{2}\). 3. Evaluate \(h(0)=\frac{-12}{4}=-3\). Thus, the y-intercept is \((0, -3)\). 4. Set the numerator equal to zero: \(3x-12=0\), so \(x=4\). This value is in the domain, so the x-intercept is \((4, 0)\).

a) Vertical asymptote: \(x=-2\); horizontal asymptote: \(y=\frac{3}{2}\) b) \((0, -3)\) c) The graph has the x-intercept \((4, 0)\).

- For equal degrees, compare the leading coefficients. - A vertical asymptote comes from a denominator zero that does not cancel. - What condition must the replacement denominator satisfy at \(x=3\)?

1. For \(g_1\), the ratio of leading coefficients is \(\frac{4}{2}=2\), so the horizontal asymptote is \(y=2\). 2. For \(g_2\), the ratio of leading coefficients is \(\frac{2}{1}=2\), so its horizontal asymptote is also \(y=2\). 3. The denominator of \(g_1\) is zero at \(2x+6=0\), so its vertical asymptote is \(x=-3\). The denominator of \(g_2\) is zero at \(x-3=0\), so its vertical asymptote is \(x=3\). Wanjiku's claim is false. 4. A replacement denominator must be zero at \(x=3\) and must not share a factor with the numerator \(4x-2\). One possible denominator is \(x-3\); another is \(2x-6\).

a) Both horizontal asymptotes are \(y=2\). b) The claim is false. The vertical asymptotes are \(x=-3\) for \(g_1\) and \(x=3\) for \(g_2\). c) One possible replacement denominator is \(x-3\).

- Use the excluded input to determine when the denominator equals zero. - A rational expression equals zero when its numerator is zero and its denominator is not zero. - Compare the roles of the numerator and denominator.

1. The excluded input is \(-5\), so the denominator must equal zero at \(x = -5\): \(-5 + k = 0\). Therefore, \(k = 5\). 2. A rational expression equals zero when its numerator is zero and its denominator is not zero. Solve \(2x - 6 = 0\): \(2x = 6\), so \(x = 3\). 3. For \(k = 5\), the denominator at \(x = 3\) is \(3 + 5 = 8\), so the function is defined there. The \(x\)-intercept is \((3, 0)\). 4. In general, the numerator determines where the function can equal zero. The denominator determines where the function is undefined. Therefore, the intercept remains at \(x = 3\) whenever \(3 + k \neq 0\).

a) \(k = 5\) b) The \(x\)-intercept is \((3, 0)\). c) The numerator determines the zero, while the denominator only determines excluded inputs. The intercept remains at \(x = 3\) as long as \(3 + k \neq 0\).

- Use the given \(x\)-intercept to set the function value equal to \(0\). - The excluded input makes the denominator zero. - Substitute \(x = 0\) to find the \(y\)-intercept.

1. The \(x\)-intercept at \(x = 5\) means \(g(5) = 0\). Substitute: \(0 = \frac{k}{5 - 3} - 4\). 2. Then \(0 = \frac{k}{2} - 4\), so \(k = 8\). 3. The denominator is zero at \(x = 3\), so the domain is \(D = \mathbb{R} \setminus \{3\}\). 4. Substitute \(x = 0\) to find the \(y\)-intercept: \(g(0) = \frac{8}{0 - 3} - 4 = -\frac{8}{3} - \frac{12}{3} = -\frac{20}{3}\). 5. Therefore, the \(y\)-intercept is \(\left(0, -\frac{20}{3}\right)\).

a) \(k = 8\) b) \(D = \mathbb{R} \setminus \{3\}\) c) The \(y\)-intercept is \(\left(0, -\frac{20}{3}\right)\).

- To find an \(x\)-intercept, what function value should you use? - Compare the equations you get for the two functions before solving. - For the domains, where does each original denominator become zero?

1. Set \(h_1(x)=0\): \(\frac{6}{x}-2=0\), so \(2x=6\) and \(x=3\). The \(x\)-intercept is \((3,0)\). 2. For \(h_2\), the equation \(\frac{6}{x-2}=0\) has no solution because the numerator is the nonzero constant \(6\). Thus the graph has no \(x\)-intercept. 3. The denominator of \(h_1\) is zero at \(x=0\), giving domain \(\mathbb{R}\setminus\{0\}\). 4. The denominator of \(h_2\) is zero at \(x=2\), giving domain \(\mathbb{R}\setminus\{2\}\).

a) The \(x\)-intercept of \(h_1\) is \((3,0)\). b) \(h_2\) has no \(x\)-intercept because its numerator is the nonzero constant \(6\). c) Domains: \(h_1\): \(\mathbb{R}\setminus\{0\}\); \(h_2\): \(\mathbb{R}\setminus\{2\}\).

- Clear the denominator when solving the intersection equation. - A contradiction means that the two graphs have no point in common. - Compare the leading coefficients of the numerator and denominator.

1. The denominator is zero when \(2x + 4 = 0\), so \(x = -2\). Therefore, the domain is \(D = \mathbb{R} \setminus \{-2\}\). 2. For the \(y\)-intercept, substitute \(x = 0\): \(g(0) = \frac{-6}{4} = -1.5\). Thus, the \(y\)-intercept is \((0, -1.5)\). 3. For the \(x\)-intercept, set the numerator equal to zero: \(3x - 6 = 0\), so \(x = 2\). Since the denominator is nonzero at \(x = 2\), the \(x\)-intercept is \((2, 0)\). 4. To test for an intersection with \(y = 1.5\), solve \(1.5 = \frac{3x - 6}{2x + 4}\). Multiplying by \(2x + 4\) gives \(1.5(2x + 4) = 3x - 6\), or \(3x + 6 = 3x - 6\). 5. This simplifies to the contradiction \(6 = -6\), so there is no intersection. The function never equals \(1.5\); the line \(y = 1.5\) is the horizontal asymptote.

a) \(D = \mathbb{R} \setminus \{-2\}\); \(x\)-intercept \((2, 0)\); \(y\)-intercept \((0, -1.5)\). b) There is no intersection. The equation leads to \(6 = -6\), so \(g(x)\) never equals \(1.5\), and \(y = 1.5\) is the horizontal asymptote.

- Check whether \(x = 0\) is allowed for each function. - A rational expression equals \(0\) only when its numerator equals \(0\). - An equation with no solution means the corresponding intercept does not exist.

1. For \(h_1\), \(x = 0\) is excluded from the domain, so its graph has no \(y\)-intercept. 2. To find an \(x\)-intercept of \(h_1\), solve \(0 = \frac{4}{x} - 1\). Then \(1 = \frac{4}{x}\), so \(x = 4\). Thus, \(h_1\) has the \(x\)-intercept \((4, 0)\). 3. For \(h_2\), substitute \(x = 0\): \(h_2(0) = \frac{4}{-1} = -4\). Thus, its \(y\)-intercept is \((0, -4)\). 4. The equation \(\frac{4}{x - 1} = 0\) has no solution because its numerator is the nonzero constant \(4\). Therefore, \(h_2\) has no \(x\)-intercept.

\(h_1\) has no \(y\)-intercept because \(x = 0\) is not in its domain; it has the \(x\)-intercept \((4, 0)\). \(h_2\) has the \(y\)-intercept \((0, -4)\) but no \(x\)-intercept because its numerator is never zero.

- The denominator is zero at a vertical asymptote. - At an \(x\)-intercept, the function value is \(0\). - Treat the entire denominator as one factor when clearing the fraction.

1. The denominator is zero at the vertical asymptote. Since \(x - b = 0\) at \(x = 2\), \(2 - b = 0\), so \(b = 2\). 2. The function is \(g(x) = \frac{4}{x - 2} + 5\). Set \(g(x) = 0\): \(0 = \frac{4}{x - 2} + 5\). 3. Then \(-5 = \frac{4}{x - 2}\), so \(-5(x - 2) = 4\). 4. Thus, \(-5x + 10 = 4\), so \(-5x = -6\) and \(x = 1.2\). 5. Therefore, the \(x\)-intercept is \((1.2, 0)\).

a) \(b = 2\) b) The \(x\)-intercept is \((1.2, 0)\).

- Which parameters are determined directly by the asymptotes? - What are the coordinates of the origin? - Substitute the asymptote values into the general form first. - How can the known point be used to solve for the remaining parameter?

1. The asymptotes give \(x_0 = 2\) and \(y_0 = -4\), so \(g(x) = \frac{a}{x - 2} - 4\). 2. Substitute the origin: \(0 = \frac{a}{0 - 2} - 4\). 3. Then \(4 = -\frac{a}{2}\), so \(a = -8\). 4. Therefore, \(g(x) = -\frac{8}{x - 2} - 4\).

\(a = -8\), and \(g(x) = -\frac{8}{x - 2} - 4\).

- What equation expresses the given y-intercept of \(h_2\)? - To compare vertical asymptotes, check where each denominator becomes zero and whether its numerator can cancel that zero. - To compare horizontal asymptotes, imagine very large positive and negative inputs.

1. Substitute \(P(0,5)\) into \(h_2\): \(5=\frac{k}{0-1}+1\), so \(5=-k+1\) and \(k=-4\). 2. For \(h_1\), the numerator is the nonzero constant \(4\). For \(h_2\), the condition \(k\neq0\) also keeps the numerator nonzero. As \(x\) approaches \(1\), the denominator approaches \(0\), so both graphs have vertical asymptote \(x=1\). 3. For very large positive or negative inputs, the fractional terms approach \(0\). Thus both graphs have horizontal asymptote \(y=1\). 4. Changing the nonzero numerator parameter scales and may reflect the graph, but does not move either asymptote.

a) \(k=-4\) b) Both graphs have vertical asymptote \(x=1\) and horizontal asymptote \(y=1\); changing \(k\neq0\) does not move them.

- What equation represents the x-axis as a horizontal line? - Where does the vertical-asymptote value appear in the function rule? - How can the y-intercept be used to find the unknown parameter?

1. Since the horizontal asymptote is the x-axis, \(y_0 = 0\). 2. The vertical asymptote \(x = 1.5\) gives \(x_0 = 1.5\), so \(f(x) = \frac{a}{x - 1.5}\). 3. Use the y-intercept: \(2 = f(0) = \frac{a}{0 - 1.5}\). 4. Therefore, \(a = 2 \cdot (-1.5) = -3\). 5. The function rule is \(f(x) = -\frac{3}{x - 1.5}\).

\(f(x) = -\frac{3}{x - 1.5}\)

- What does the excluded domain value reveal about the vertical asymptote? - What equation represents an x-intercept? - How is a function value found for a given input?

1. The excluded domain value gives the vertical asymptote \(x = 1\), so \(x_0 = 1\). The horizontal asymptote gives \(y_0 = -2\). 2. Write \(f(x) = \frac{a}{x - 1} - 2\). 3. The x-intercept at \(x = 5\) means \(f(5) = 0\): \(0 = \frac{a}{5 - 1} - 2\). Then \(2 = \frac{a}{4}\), so \(a = 8\). 4. Thus, \(f(x) = \frac{8}{x - 1} - 2\). 5. Evaluate \(f(3) = \frac{8}{3 - 1} - 2 = 2\).

a) \(f(x) = \frac{8}{x - 1} - 2\) b) \(y = 2\)

- Check whether \(x = 0\) is in the domain of \(f\). - A rational expression equals \(0\) only when its numerator equals \(0\). - For \(g\), use the usual conditions for the two intercepts.

1. A \(y\)-intercept of \(f\) would require \(x = 0\), but \(f\) is undefined there. An \(x\)-intercept would require \(\frac{1}{x} = 0\), which has no solution because the numerator is the nonzero constant \(1\). Therefore, \(f\) has no intercepts. 2. For the \(y\)-intercept of \(g\), substitute \(x = 0\): \(g(0) = \frac{4}{-2} + 1 = -1\). Thus, the \(y\)-intercept is \((0, -1)\). 3. For the \(x\)-intercept of \(g\), solve \(0 = \frac{4}{x - 2} + 1\). Then \(-1 = \frac{4}{x - 2}\), so \(x - 2 = -4\) and \(x = -2\). Thus, the \(x\)-intercept is \((-2, 0)\).

a) The graph of \(f\) has no \(y\)-intercept because \(x = 0\) is excluded, and no \(x\)-intercept because its numerator is never zero. b) The intercepts of \(g\) are \((-2, 0)\) and \((0, -1)\).

- Find where the denominator is zero and what value the function approaches for very large inputs. - What part of the formula measures the vertical distance from the horizontal asymptote? - Compare each distance with the strip's half-height.

1. The denominator is zero at \(x=1\), giving the vertical asymptote \(x=1\). For very large positive or negative inputs, the fraction approaches \(0\), so the horizontal asymptote is \(y=1.5\). 2. A point is inside the strip if its vertical distance from \(y=1.5\) is less than \(0.4\). 3. At \(x=3\), \(f(3)=1.5-\frac{2}{2}=0.5\), a distance of \(1\) from the asymptote. That point is outside. 4. At \(x=7\), \(f(7)=1.5-\frac{2}{6}=\frac{7}{6}\). Its distance from \(1.5\) is \(\frac{1}{3}<0.4\), so that point is inside.

a) Vertical asymptote: \(x=1\); horizontal asymptote: \(y=1.5\) b) At \(x=3\): outside. At \(x=7\): inside.

- Find where the denominator is zero and what output the function approaches at infinity. - Express vertical distance from the horizontal asymptote with an absolute value. - Use the condition \(x>-2\) to determine the sign of the denominator.

1. The denominator is zero at \(0.5x+1=0\), so the vertical asymptote is \(x=-2\). As \(|x|\to\infty\), the fraction approaches \(0\), so the horizontal asymptote is \(y=-2\). 2. The required condition is \(|g(x)+2|<0.1\), or \(\left|\frac{3}{0.5x+1}\right|<0.1\). 3. For \(x>-2\), the denominator is positive, so \(\frac{3}{0.5x+1}<0.1\). 4. Multiply by the positive denominator: \(3<0.1(0.5x+1)=0.05x+0.1\). Thus, \(2.9<0.05x\), which gives \(x>58\).

a) Vertical asymptote: \(x=-2\); horizontal asymptote: \(y=-2\) b) \(x>58\)

- Compare the rule with \(y=\frac{a}{x-h}+k\). - Apply the odd-function condition. - Consider what remains as \(x\to\infty\).

1. A graph of the form \(y=\frac{a}{x-h}+k\) has center of symmetry \((h, k)\). Since \(x+b=x-(-b)\), the center is \((-b, c)\). 2. Origin symmetry would require \(f(-x)=-f(x)\). As \(x\to\infty\), both fractional terms in this equation approach \(0\), leaving the necessary condition \(c=-c\). 3. That condition implies \(c=0\). Therefore, when \(c\ne0\), the graph cannot be symmetric about the origin.

1) The center of symmetry is \((-b, c)\). 2) Origin symmetry requires \(c=0\), so it is impossible when \(c\ne0\).

- Build the rule from even powers of \(x\). - Use the ratio of leading coefficients for the horizontal asymptote. - Substitute \(x=0\) to impose the y-intercept condition.

1. To make the function even, use only even powers of \(x\). Begin with \(f(x)=\frac{ax^2+b}{x^2+c}\). 2. Equal numerator and denominator degrees give a horizontal asymptote equal to the ratio of leading coefficients. To obtain \(y=3\), choose \(a=3\). 3. The y-intercept condition requires \(f(0)=\frac{b}{c}=1\), so choose equal nonzero values for \(b\) and \(c\), such as \(b=c=1\). 4. This gives \(f(x)=\frac{3x^2+1}{x^2+1}\), which is defined for every real \(x\), is even, has y-intercept \((0, 1)\), and has horizontal asymptote \(y=3\).

One possible rule is \(f(x)=\frac{3x^2+1}{x^2+1}\).

- Which power of \(x\) is highest in both polynomials? - What expression remains from the \(x^2\) term after you divide by \(x^3\)? - As the input becomes very large and negative, what happens to each reciprocal-power term?

1. For \(x\ne0\), write the numerator as \(x^3\left(\frac{4}{x^3}-2\right)\). 2. Write the denominator as \(x^3\left(0.5+\frac{1}{x}-\frac{1}{x^3}\right)\). 3. Cancel \(x^3\), giving \(g(x)=\frac{\frac{4}{x^3}-2}{0.5+\frac{1}{x}-\frac{1}{x^3}}\) for \(x\ne0\). 4. For very large negative inputs, the terms involving reciprocal powers approach \(0\). The outputs approach \(\frac{-2}{0.5}=-4\), so the horizontal asymptote is \(y=-4\).

a) \(g(x)=\frac{\frac{4}{x^3}-2}{0.5+\frac{1}{x}-\frac{1}{x^3}}\) for \(x\ne0\) b) The outputs approach \(-4\); horizontal asymptote: \(y=-4\).

- What changes in the highest power of the numerator when \(a=0\)? - For equal-degree polynomials, compare their leading terms. - For part b, compare the degrees again after substituting the parameter.

1. When \(a\ne0\), both polynomials have degree \(2\). For very large inputs, their leading terms dominate, so the fraction approaches \(\frac{a}{3}\). 2. Set \(\frac{a}{3}=4\), giving \(a=12\). 3. When \(a=0\), the function becomes \(f_0(x)=\frac{6x}{3x^2-12}\). The denominator has higher degree, so the outputs approach \(0\) at both ends. 4. The horizontal asymptote in part b is \(y=0\).

a) \(a=12\) b) The function values approach \(0\) at both ends; horizontal asymptote: \(y=0\).

- Try factoring both polynomials before deciding where the graph has an asymptote. - What does a factor canceled from both numerator and denominator leave behind? - For very large inputs, what happens to a fixed numerator divided by a growing denominator?

1. Factor the expression: \(g(x)=\frac{3(x-2)}{(x-2)(x+2)}\). 2. For \(x\ne2\), this simplifies to \(g(x)=\frac{3}{x+2}\). The original domain excludes both \(x=-2\) and \(x=2\). 3. The canceled factor leaves a hole at \(x=2\), not a vertical asymptote. The uncanceled denominator factor gives a vertical asymptote at \(x=-2\). 4. For very large positive or negative inputs, \(\frac{3}{x+2}\) approaches \(0\). The horizontal asymptote is \(y=0\).

Vertical asymptote: \(x=-2\) Horizontal asymptote: \(y=0\)

- First find every value that makes the denominator zero. - Check whether the numerator is also zero at any of those values, and simplify if possible. - Use the multiplicity of each denominator factor after all common factors have been canceled. - What does an odd or even multiplicity tell you about the sign of the function on opposite sides of a vertical asymptote?

1. The denominator is zero at \(x=1\), with multiplicity \(1\), and at \(x=-2\), with multiplicity \(2\). 2. Factor the numerator: \(2x+4=2(x+2)\). The numerator and denominator share one factor of \(x+2\). 3. Simplify the expression for \(x\neq -2\): \(g(x)=\frac{2(x+2)}{(x-1)(x+2)^2}=\frac{2}{(x-1)(x+2)}\). 4. The simplified denominator has zeros at \(x=1\) and \(x=-2\), each with multiplicity \(1\), so both values are vertical asymptotes. 5. Each remaining denominator zero has odd multiplicity. Therefore, the graph changes sign across both vertical asymptotes.

The vertical asymptotes are \(x=1\) and \(x=-2\). The graph changes sign across both asymptotes because the factors \(x-1\) and \(x+2\) each have odd multiplicity \(1\) in the simplified expression.

- Substitute each server count into the model and keep the dollar units. - To find the least count under the cost threshold, write an inequality and check which side of zero the denominator lies on. - How should a continuous inequality answer be adjusted for a whole number of servers? - For long-term behavior, compare the highest-degree terms and interpret the result in dollars per server.

1. Evaluate and round to cents: \(B(20)=\frac{6000}{17}\approx\$352.94\), \(B(100)=\frac{20400}{65}\approx\$313.85\), and \(B(500)=\frac{92400}{305}\approx\$302.95\). The cost per server decreases with the server count. 2. Solve \(\frac{180x+2400}{0.6x+5}<320\). The denominator is positive for \(x\ge10\), so \(180x+2400<192x+1600\), giving \(x>\frac{200}{3}\). 3. The least whole-number server count that works is \(67\). 4. For very large server counts, the leading terms dominate: \(B(x)\) approaches \(\frac{180}{0.6}=300\). The horizontal asymptote is \(y=300\), measured in dollars per server. Since \(B(x)-300=\frac{900}{0.6x+5}>0\), the modeled cost approaches \(\$300\) from above.

a) \(B(20)\approx\$352.94\), \(B(100)\approx\$313.85\), and \(B(500)\approx\$302.95\); the cost decreases. b) \(67\) servers c) The cost approaches \(\$300\) per server from above; horizontal asymptote: \(y=300\).

- Track the units: each output represents hundreds of animals. - What input represents the beginning of the observations? - Convert the target animal count into the units used by the function before solving. - For large values of \(t\), compare the highest-degree terms of the numerator and denominator.

1. \(N(0)=\frac{50}{5}=10\) hundreds, so the initial population is \(1000\) animals. 2. A population of \(5500\) animals means \(N(t)=55\). Solve \(\frac{120t+50}{2t+5}=55\), giving \(120t+50=110t+275\), so \(t=22.5\) years. 3. For very large positive \(t\), the leading terms dominate and the output approaches \(\frac{120}{2}=60\) hundreds. The horizontal asymptote is \(y=60\), representing a long-term population of about \(6000\) animals in the model.

a) \(1000\) animals b) \(22.5\) years c) The modeled population approaches \(6000\) animals; horizontal asymptote: \(y=60\) (hundreds of animals).

- Begin by finding the zeros of each denominator. - Check for factors shared with the numerator before deciding what each excluded input produces. - Compare the signs of the expression just to the left and right of each remaining denominator zero. - How does an even or odd denominator multiplicity affect those signs?

1. For part a, the denominator has a double zero at \(x=1\), where the numerator is \(5\). Thus \(x=1\) is a vertical asymptote; the even denominator multiplicity means the graph does not change sign across it. 2. For part b, \(x^2-4x=x(x-4)\). Both denominator zeros, \(x=0\) and \(x=4\), are simple and the numerator stays nonzero. Both are vertical asymptotes, and the graph changes sign across each. 3. For part c, factor \(x^2-4=(x-2)(x+2)\). Canceling \(x-2\) gives \(h(x)=\frac{1}{x+2}\) for \(x\ne2\). Thus \(x=2\) is a hole, not an asymptote. The remaining simple denominator zero at \(x=-2\) gives a vertical asymptote where the graph changes sign.

a) Vertical asymptote: \(x=1\); no sign change across it. b) Vertical asymptotes: \(x=0\) and \(x=4\); the graph changes sign across both. c) Vertical asymptote: \(x=-2\), with a sign change. There is a hole at \(x=2\).

- Find the excluded inputs by factoring the original denominator. - Which denominator factors, if any, can be canceled? - Look at the sign of the simplified function just to either side of the remaining denominator zero.

1. Factor: \(x^2-x-6=(x-3)(x+2)\) and \(2x-6=2(x-3)\). The excluded inputs are \(x=3\) and \(x=-2\). 2. For \(x\ne3\), simplify to \(f(x)=\frac{2}{x+2}\). The canceled factor leaves a hole at \(x=3\). 3. The remaining factor \(x+2\) has a simple zero at \(x=-2\), giving a vertical asymptote. The graph changes sign across that line because the denominator changes sign while the numerator stays positive.

There is a hole at \(x=3\) and a vertical asymptote at \(x=-2\). The graph changes sign across \(x=-2\).

- Simplify the rational expression before deciding how the graph behaves near the excluded value. - How does a common factor in the numerator and denominator affect the remaining multiplicity? - Which multiplicities produce a sign change across a vertical asymptote?

1. Factor the denominator: \(x^2-4x+4=(x-2)^2\). The numerator also contains the factor \(x-2\). 2. Simplify for \(x\neq 2\): \(f(x)=\frac{x-2}{(x-2)^2}=\frac{1}{x-2}\). 3. In the simplified expression, the denominator factor \(x-2\) has multiplicity \(1\), which is odd. Therefore, \(x=2\) is a vertical asymptote and the graph changes sign across it. 4. Lara’s conclusion is incorrect because she used the multiplicity in the original denominator without first canceling the common factor.

Lara’s reasoning is incorrect. Since \(x^2-4x+4=(x-2)^2\), the expression simplifies to \(f(x)=\frac{1}{x-2}\) for \(x\neq 2\). The remaining denominator factor has odd multiplicity \(1\), so \(x=2\) is a vertical asymptote and the graph changes sign across it.

- Factor the numerator before deciding what happens at the excluded input. - Does the common factor cancel completely, or does a denominator factor remain? - Compare the signs on both sides of \(x=3\) using the simplified expression.

1. Factor the numerator as \(x^2-9=(x-3)(x+3)\). 2. The denominator is zero at \(x=3\), so the original function is undefined there. 3. For \(x\ne3\), simplify to \(k(x)=\frac{x+3}{(x-3)^2}\). 4. Near \(x=3\), the numerator is positive and the squared denominator is positive on both sides. The denominator approaches zero, so the graph rises without bound on both sides. It does not change sign across the vertical asymptote \(x=3\).

The line \(x=3\) is a vertical asymptote. The graph rises on both sides and does not change sign across it.

- Begin by finding the real zeros of each denominator. - Check whether any common numerator and denominator factors cancel. - After simplifying, compare signs just to the left and right of each remaining denominator zero. - What happens when a denominator is positive for every real input?

1. For part a, factor to obtain \(\frac{3(x-4)}{(x-4)(x+4)}=\frac{3}{x+4}\) for \(x\ne4\). There is a hole at \(x=4\). The remaining simple denominator zero at \(x=-4\) gives a vertical asymptote where the sign changes. 2. For part b, factor \(x^2+4x+4=(x+2)^2\). For \(x\ne-2\), simplify to \(\frac{1}{x+2}\). The line \(x=-2\) is a vertical asymptote where the sign changes. 3. For part c, \(x^2+1>0\) for every real \(x\), so the function has no excluded real inputs, holes, or vertical asymptotes.

a) Vertical asymptote: \(x=-4\), with a sign change; hole at \(x=4\). b) Vertical asymptote: \(x=-2\), with a sign change. c) No vertical asymptotes or holes.

- Average cost equals total cost divided by the number of items. - Decide which values of \(x\) are possible in the context. - What happens to \(\frac{1200}{x}\) as \(x\) increases? - Use an inequality to compare the average cost with \(\$4.00\).

1. The total cost is \(K(x)=3.50x+1200\). 2. The average cost is \(k(x)=\frac{K(x)}{x}=3.50+\frac{1200}{x}\), with \(x\in\mathbb{N}\) and \(x\ge 1\). 3. Since \(\frac{1200}{x}\to 0\) as \(x\to\infty\), \(k(x)\to 3.50\). The horizontal asymptote is \(y=3.50\). For large production levels, the average cost approaches the variable cost per case. 4. Solve \(3.50+\frac{1200}{x}<4.00\). For \(x>0\), this gives \(\frac{1200}{x}<0.50\), so \(x>2400\). 5. Therefore, the least whole-number production level is \(2401\) cases.

a) \(k(x)=\frac{3.50x+1200}{x}=3.50+\frac{1200}{x}\), where \(x\in\mathbb{N}\) and \(x\ge 1\) b) The horizontal asymptote is \(y=3.50\). The average cost approaches \(\$3.50\) per case as production increases. c) \(2401\) cases

- For which values of \(x\) is the denominator zero? - How does the sign of \(k\) affect the equation \(x^2=2k\)? - A denominator zero gives a vertical asymptote when the numerator is nonzero there. - How many real solutions can \(x^2=c\) have, depending on the value of \(c\)?

1. If \(k<0\), then \(2k<0\), so \(x^2-2k=x^2+(-2k)>0\) for every real \(x\). Thus, \(D_{f_k}=\mathbb{R}\). 2. If \(k=0\), the denominator is \(x^2\), which is zero only at \(x=0\). Thus, \(D_{f_0}=\mathbb{R}\setminus\{0\}\). 3. If \(k>0\), solving \(x^2-2k=0\) gives \(x=\pm\sqrt{2k}\). Thus, \(D_{f_k}=\mathbb{R}\setminus\{-\sqrt{2k}, \sqrt{2k}\}\). 4. For \(k=8\), \(f_8(x)=\frac{6}{x^2-16}\). The denominator is zero at \(x=-4\) and \(x=4\), while the numerator is nonzero. Therefore, the vertical asymptotes are \(x=-4\) and \(x=4\). 5. There are no vertical asymptotes when \(k<0\), one when \(k=0\), and two when \(k>0\).

a) If \(k<0\), \(D_{f_k}=\mathbb{R}\). If \(k=0\), \(D_{f_0}=\mathbb{R}\setminus\{0\}\). If \(k>0\), \(D_{f_k}=\mathbb{R}\setminus\{-\sqrt{2k}, \sqrt{2k}\}\). b) \(x=-4\) and \(x=4\) c) \(0\) vertical asymptotes for \(k<0\), \(1\) for \(k=0\), and \(2\) for \(k>0\).

- Factor and simplify each fraction before making a graph prediction. - Which denominator factors remain after cancellation? - Compare the sign of each simplified expression just to the left and right of the excluded input.

1. For part a, factor and simplify: \(f(x)=\frac{2(x-2)}{(x-2)^3}=\frac{2}{(x-2)^2}\) for \(x\ne2\). The remaining denominator zero at \(x=2\) has even multiplicity, so \(x=2\) is a vertical asymptote with no sign change. 2. For part b, \(g(x)=\frac{(x-1)(x+1)}{(x+1)^2}=\frac{x-1}{x+1}\) for \(x\ne-1\). The remaining denominator zero at \(x=-1\) is simple, so the graph changes sign across this vertical asymptote. 3. Neither excluded input is a hole because each still makes the simplified denominator zero.

a) Vertical asymptote: \(x=2\); no sign change. b) Vertical asymptote: \(x=-1\); the graph changes sign. Neither function has a hole.

- Factor the numerator and denominator separately. - Which excluded inputs disappear from the denominator after cancellation? - At a remaining denominator zero, compare signs on the two sides.

1. Factor the denominator as \((x-3)(x+3)\), so the domain is \(\mathbb{R}\setminus\{-3,3\}\). 2. Factor the numerator as \((x-3)(x+1)\). For \(x\ne3\), simplify to \(h(x)=\frac{x+1}{x+3}\). 3. Because the factor \(x-3\) cancels completely, there is a hole at \(x=3\). 4. The remaining simple denominator zero at \(x=-3\) gives a vertical asymptote. Near \(x=-3\), the numerator stays negative but the denominator changes sign, so the graph changes sign across it.

Domain: \(\mathbb{R}\setminus\{-3,3\}\) Hole: \(x=3\) Vertical asymptote: \(x=-3\), with a sign change.

- Compare the degrees of each numerator and denominator. - What happens to the ratio when the denominator's degree is greater? - For equal degrees, which pair of coefficients controls the far-end behavior?

1. For \(f_2\) and \(f_4\), the numerator has lower degree than the denominator, so both outputs approach \(0\) at both ends. Their horizontal asymptote is \(y=0\). 2. For \(f_1\), \(f_3\), and \(f_5\), the numerator and denominator have equal degree. Their leading coefficient ratios are \(\frac{3}{1}=3\), \(\frac{6}{2}=3\), and \(\frac{-9}{-3}=3\). 3. Thus \(f_1\), \(f_3\), and \(f_5\) approach \(3\) at both ends, with horizontal asymptote \(y=3\).

Horizontal asymptote \(y=0\): \(f_2\) and \(f_4\) Horizontal asymptote \(y=3\): \(f_1\), \(f_3\), and \(f_5\)

- Start by factoring both the numerator and denominator. - Which denominator zeros remain after you cancel common factors? - If every factor of the denominator cancels, does the original expression become defined at those excluded inputs?

1. For part a, factor the numerator as \((x-3)(x+3)\) and the denominator as \((x-3)(x+1)\). The domain excludes \(x=3\) and \(x=-1\). After canceling, \(f(x)=\frac{x+3}{x+1}\) for \(x\ne3\), so there is a hole at \(x=3\) and a vertical asymptote at \(x=-1\). 2. For part b, factor \(x^3-4x=x(x-2)(x+2)\) and \(x^2-2x=x(x-2)\). The domain excludes \(x=0\) and \(x=2\). Both factors cancel, leaving \(g(x)=x+2\) away from those inputs. Both excluded inputs give holes and there is no vertical asymptote.

a) Excluded inputs: \(x=-1\) and \(x=3\); vertical asymptote at \(x=-1\), hole at \(x=3\). b) Excluded inputs: \(x=0\) and \(x=2\); holes at both; no vertical asymptote.

- Determine the excluded inputs from the original denominator. - Factor and cancel before deciding which excluded inputs are holes. - For any factor still in the denominator, inspect the sign on either side of its zero.

1. For part a, factor \(h(x)=\frac{x(x-2)}{(x-2)^2}\). The domain excludes \(x=2\). After canceling one factor, \(h(x)=\frac{x}{x-2}\) for \(x\ne2\). The remaining simple denominator zero is a vertical asymptote at \(x=2\), where the graph changes sign. 2. For part b, factor \(k(x)=\frac{(x+3)(x-2)}{x(x+3)}\). The domain excludes \(x=-3\) and \(x=0\). After cancellation, \(k(x)=\frac{x-2}{x}\) for \(x\ne-3\). There is a hole at \(x=-3\), and the remaining simple zero at \(x=0\) gives a vertical asymptote where the graph changes sign.

a) Domain: \(\mathbb{R}\setminus\{2\}\); vertical asymptote: \(x=2\), with a sign change; no hole. b) Domain: \(\mathbb{R}\setminus\{-3,0\}\); hole at \(x=-3\); vertical asymptote: \(x=0\), with a sign change.

- Begin by factoring both numerators and denominators. - Which excluded inputs disappear from the denominator after cancellation? - Check each excluded input separately rather than assuming all denominator zeros give the same feature.

1. Factor \(f\): its denominator is \(x(x-3)\), and its numerator is \(x(x-3)(x+1)\). Both denominator factors cancel, leaving \(f(x)=x+1\) for \(x\ne0,3\). Thus the graph has holes at \(x=0\) and \(x=3\). 2. Factor \(g\): its denominator is \((x-2)(x+2)\), while its numerator is \(0.2(x-2)^2\). Canceling one factor gives \(g(x)=\frac{0.2(x-2)}{x+2}\) for \(x\ne2\). The canceled factor leaves a hole at \(x=2\). The other excluded value, \(x=-2\), is a vertical asymptote, not a hole.

Holes of \(f\): \(x=0\) and \(x=3\). Hole of \(g\): \(x=2\).

- When is a rational expression undefined? - Try writing the numerator as a multiple of the denominator plus a remainder. - When does a rational expression equal \(0\)? - Set the function equal to \(1\) and solve for \(x\). - What happens to a fraction with fixed numerator as its denominator grows?

1. The denominator cannot equal \(0\), so \(x \ne -2\). The domain is \(\mathbb{R} \setminus \{-2\}\). 2. Rewrite the numerator: \(3x - 2 = 3(x + 2) - 8\). Thus, \(\frac{3x - 2}{x + 2} = \frac{3(x + 2) - 8}{x + 2} = 3 - \frac{8}{x + 2}\). 3. A rational expression is \(0\) when its numerator is \(0\) and its denominator is nonzero. Solve \(3x - 2 = 0\), giving \(x = \frac{2}{3}\). 4. Solve \(1 = 3 - \frac{8}{x + 2}\). Then \(2 = \frac{8}{x + 2}\), so \(2(x + 2) = 8\) and \(x = 2\). 5. As \(x \to \infty\), \(\frac{8}{x + 2} \to 0\), so \(g(x) \to 3\). The function never equals \(3\) because \(\frac{8}{x + 2}\) cannot equal \(0\).

1) \(\mathbb{R} \setminus \{-2\}\) 2) \(g(x) = \frac{3(x + 2) - 8}{x + 2} = 3 - \frac{8}{x + 2}\) 3) \(x = \frac{2}{3}\) 4) \(x = 2\) 5) The function values approach \(3\), but the function never attains \(3\).

- Rewrite negative exponents as reciprocals. - Determine the sign of each function for positive and negative inputs. - Consider what happens to a fraction when its denominator becomes very large or approaches zero. - Set the functions equal to find the intersection.

1. Both functions are undefined at \(x=0\), so each domain is \(\mathbb{R}\setminus\{0\}\). The graph of \(f(x)=\frac{1}{x}\) lies in Quadrants I and III, while the graph of \(g(x)=\frac{1}{x^2}\) lies in Quadrants I and II. 2. \(f(0.1)=10\) and \(g(0.1)=100\). Since \(x^{-2}\) has the greater order of reciprocal growth, \(g(x)\) approaches positive infinity faster as \(x\to0^+\). 3. As \(x\to\infty\), both function values approach \(0\). Both graphs approach the horizontal asymptote \(y=0\). 4. Set the functions equal: \(x^{-1}=x^{-2}\). Multiplying by \(x^2\) gives \(x=1\), and the corresponding y-value is \(1\). The intersection is \((1, 1)\).

a) Both domains are \(\mathbb{R}\setminus\{0\}\). The graph of \(f\) lies in Quadrants I and III; the graph of \(g\) lies in Quadrants I and II. b) \(f(0.1)=10\) and \(g(0.1)=100\); \(g\) approaches positive infinity faster as \(x\to0^+\). c) Both approach \(0\), so the horizontal asymptote is \(y=0\). d) \((1, 1)\)

- For the domain, compare the cases \(a>0\) and \(a<0\). - To test evenness, substitute \(-x\) into the rule and compare it with the original expression. - Find zeros from the numerator, but check whether those inputs are allowed. - Compare the leading terms for large positive or negative inputs.

1. For \(a>0\), \(x^2+a>0\) for every real \(x\), giving full domain. For \(a<0\), the denominator is zero at \(x=\pm\sqrt{-a}\). Thus statement 1 depends on \(a\) and is true exactly when \(a>0\). 2. Since \(g_a(-x)=\frac{(-x)^2-a}{(-x)^2+a}=g_a(x)\), every graph is symmetric about the y-axis. Statement 2 is always true. 3. Numerator zeros satisfy \(x^2=a\), giving two real zeros for \(a>0\) and none for \(a<0\). When \(a>0\), neither zero is excluded. Statement 3 depends on \(a\), being true exactly when \(a>0\). 4. For very large positive or negative inputs, the leading terms \(x^2\) in numerator and denominator dominate, so the output approaches \(1\). Statement 4 is always true.

1) Depends on \(a\); true exactly when \(a>0\). 2) True for every \(a\ne0\). 3) Depends on \(a\); true exactly when \(a>0\). 4) True for every \(a\ne0\).

- Decide whether \(x^2+1\) can equal \(0\) for a real value of \(x\). - Set the function equal to \(0\) to find the x-intercepts. - Clear the denominator and solve the resulting equation. - Check the solutions against the domain.

1. Since \(x^2+1>0\) for every real \(x\), the denominator is never zero. Therefore, the domain is \(\mathbb{R}\), and there are no vertical asymptotes. 2. To find the x-intercepts, set \(f(x)=0\): \(\frac{x^2-5}{x^2+1}+1=0\). 3. Then \(\frac{x^2-5}{x^2+1}=-1\), so \(x^2-5=-(x^2+1)\). 4. Solving gives \(2x^2=4\), so \(x^2=2\) and \(x=\pm\sqrt{2}\). 5. Both values are in the domain, so the x-intercepts are \((-\sqrt{2}, 0)\) and \((\sqrt{2}, 0)\).

Domain: \(\mathbb{R}\); no vertical asymptotes; x-intercepts: \((-\sqrt{2}, 0)\) and \((\sqrt{2}, 0)\)

- Factor both polynomials and identify the original excluded inputs. - Which factors remain after simplifying the fraction? - To locate a hole, use the simplified expression at the excluded input. - Check each numerator zero against the original domain before calling it an x-intercept.

1. Factor the expression: \(f(x)=\frac{(x-2)(x+1)}{(x-2)(x+2)}\). The original denominator is zero at \(x=-2\) and \(x=2\), so the domain is \(\mathbb{R}\setminus\{-2,2\}\). 2. On its domain, \(f(x)=\frac{x+1}{x+2}\). The remaining denominator factor gives a vertical asymptote at \(x=-2\). 3. The canceled factor leaves a hole at \(x=2\). Substitute into the simplified expression to find its y-coordinate: \(\frac{2+1}{2+2}=\frac{3}{4}\). The hole is \(\left(2,\frac{3}{4}\right)\). 4. The original numerator is zero at \(x=2\) and \(x=-1\), but only \(x=-1\) is in the domain. Thus the only x-intercept is \((-1,0)\).

Domain: \(\mathbb{R}\setminus\{-2,2\}\) Vertical asymptote: \(x=-2\) Hole: \(\left(2,\frac{3}{4}\right)\) x-intercept: \((-1,0)\)

- Factor the denominator. - A numerator zero is an x-intercept only when it lies in the domain. - Determine when a numerator factor cancels one of the denominator factors.

1. Factor the denominator: \(x^2-x-6=(x-3)(x+2)\). Therefore, every function has domain \(\mathbb{R}\setminus\{-2, 3\}\). 2. The numerator is zero at \(x=a\). If \(a\neq-2\) and \(a\neq 3\), this value is in the domain, so the x-intercept is \((a, 0)\). 3. If \(a=-2\) or \(a=3\), the numerator zero occurs at an excluded value, so the graph has no x-intercept. 4. Ordinarily, the denominator factors produce vertical asymptotes at \(x=-2\) and \(x=3\). Exactly one remains when the numerator cancels one denominator factor. 5. For \(a=3\), the factor \(x-3\) cancels, leaving only the vertical asymptote \(x=-2\). For \(a=-2\), the factor \(x+2\) cancels, leaving only the vertical asymptote \(x=3\).

a) \(\mathbb{R}\setminus\{-2, 3\}\) b) If \(a\notin\{-2, 3\}\), the x-intercept is \((a, 0)\). If \(a=-2\) or \(a=3\), there is no x-intercept. c) Exactly one vertical asymptote occurs when \(a=-2\) or \(a=3\).

- Factor out the nonzero parameter from the denominator. - Where can the numerator equal zero without excluding the input? - Identify which denominator zeros cannot cancel and compare the numerator's and denominator's degrees.

1. Factor the denominator: \(kx^2-4k=k(x-2)(x+2)\). Since \(k\ne0\), the domain is \(\mathbb{R}\setminus\{-2,2\}\), independent of \(k\). 2. The numerator is zero when \(2x+6=0\), so \(x=-3\). It is in the domain, giving x-intercept \((-3,0)\). 3. At \(x=-2\) and \(x=2\), the numerator is nonzero, so both are vertical asymptotes for every allowed \(k\). 4. The denominator has higher degree than the numerator. For very large positive or negative inputs, the quotient approaches \(0\), so the horizontal asymptote is \(y=0\) for every \(k\ne0\).

a) Domain: \(\mathbb{R}\setminus\{-2,2\}\) for every \(k\ne0\) b) x-intercept: \((-3,0)\) c) Vertical asymptotes: \(x=-2\), \(x=2\); horizontal asymptote: \(y=0\)

- Factor the denominator completely. - Examine the exponent on each denominator factor. - How do odd and even multiplicities affect sign changes across a vertical asymptote? - Check whether the numerator is also zero at either denominator zero.

1. Factor the denominator: \(x^3-6x^2+9x=x(x^2-6x+9)=x(x-3)^2\). 2. The denominator zeros are \(x=0\) and \(x=3\). 3. The numerator is \(5\) at \(x=0\) and \(8\) at \(x=3\), so neither denominator factor cancels. Both values are vertical asymptotes. 4. The factor \(x\) has odd multiplicity \(1\), so the graph changes sign across \(x=0\). 5. The factor \(x-3\) has even multiplicity \(2\), so the graph does not change sign across \(x=3\).

\(x=0\) is a vertical asymptote where the graph changes sign. \(x=3\) is a vertical asymptote where the graph does not change sign.

- Factor the difference of squares inside the denominator. - How does the outer exponent affect each factor? - What does an odd multiplicity imply about the graph near a vertical asymptote?

1. Set the denominator equal to zero: \((x^2-1)^3=0\), so \(x^2-1=0\). The solutions are \(x=-1\) and \(x=1\). 2. Factor the denominator: \((x^2-1)^3=[(x-1)(x+1)]^3=(x-1)^3(x+1)^3\). 3. The numerator is the nonzero constant \(1\), so both denominator zeros produce vertical asymptotes. 4. Each denominator factor has odd multiplicity \(3\). Therefore, the graph changes sign across both vertical asymptotes.

The vertical asymptotes are \(x=-1\) and \(x=1\). The graph changes sign across both.

- Compare the degrees and leading coefficients to find the horizontal asymptote. - Write an absolute-value expression for the vertical distance from the graph to the asymptote. - Simplify the difference before solving the inequality. - A fraction has small absolute value when its denominator has sufficiently large absolute value.

1. The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients: \(y=\frac{1}{2}\). 2. The distance from the asymptote is \(\left\lvert f(x)-\frac{1}{2}\right\rvert=\left\lvert\frac{x^2+4}{2x^2-8}-\frac{x^2-4}{2x^2-8}\right\rvert=\left\lvert\frac{4}{x^2-4}\right\rvert\). 3. Solve \(\left\lvert\frac{4}{x^2-4}\right\rvert<0.01\). This is equivalent to \(\lvert x^2-4\rvert>400\). 4. The inequality \(x^2-4>400\) gives \(x^2>404\), so \(\lvert x\rvert>\sqrt{404}\). The other case, \(x^2-4<-400\), has no real solutions. 5. Therefore, \(x<-\sqrt{404}\) or \(x>\sqrt{404}\), where \(\sqrt{404}\approx 20.10\).

1) Horizontal asymptote: \(y=\frac{1}{2}\) 2) \(x\in(-\infty, -\sqrt{404})\cup(\sqrt{404}, \infty)\), approximately \(x<-20.10\) or \(x>20.10\)

- First find the inputs that are excluded by the denominator. - The equation for an x-intercept may have no, one, or two real solutions depending on \(a\). - Check whether any possible x-intercept is an excluded input. - What changes if the numerator and denominator share a factor?

1. The denominator is zero at \(x=-3\) and \(x=3\), so both inputs are excluded for every \(a\). 2. The numerator is zero when \(x^2=a\). If \(a<0\), there are no real solutions. If \(a=0\), the only solution is \(x=0\). If \(a>0\), the possible zeros are \(x=\pm\sqrt{a}\). 3. When \(a=9\), the possible zeros are the excluded inputs \(\pm3\), so there are no x-intercepts. For \(a>0\) with \(a\ne9\), both zeros are in the domain. 4. When \(a=9\), \(f_9(x)=\frac{9-x^2}{x^2-9}=-1\) for all allowed \(x\). The graph has holes at \((-3,-1)\) and \((3,-1)\). 5. When \(a\ne9\), the numerator is nonzero at both excluded inputs, so \(x=-3\) and \(x=3\) are vertical asymptotes.

X-intercepts: none if \(a<0\) or \(a=9\); \((0,0)\) if \(a=0\); \((-\sqrt{a},0)\) and \((\sqrt{a},0)\) if \(a>0\) and \(a\ne9\). At \(x=\pm3\): vertical asymptotes if \(a\ne9\); holes at \((-3,-1)\) and \((3,-1)\) if \(a=9\).

- Where does the graph form a vertical barrier? - Which horizontal value does the graph approach from above and below? - What happens to a fraction when its denominator has a large absolute value?

1. The graph approaches the vertical line \(x = 2\), so the excluded input is \(x = 2\) and \(d = 2\). 2. The graph approaches the horizontal line \(y = 1\), so \(e = 1\). Thus, \(f(x) = \frac{2}{x - 2} + 1\). 3. Evaluate \(f(102) = \frac{2}{102 - 2} + 1 = 1.02\). 4. Evaluate \(f(-98) = \frac{2}{-98 - 2} + 1 = 0.98\). 5. Both values are close to \(1\). As \(|x|\) becomes large, \(\frac{2}{x - 2}\) approaches \(0\), so the function values approach the horizontal asymptote \(y = 1\).

a) Vertical asymptote and excluded input: \(x = 2\); \(d = 2\) b) Horizontal asymptote: \(y = 1\); \(e = 1\) c) \(f(102) = 1.02\) and \(f(-98) = 0.98\). Both are close to the horizontal asymptote \(y = 1\).

- Locate the x-value that the graph approaches but never reaches. - Identify the horizontal line approached as \(|x|\) becomes large. - Use the asymptotes to determine the horizontal and vertical translations.

1. From the graph, the vertical asymptote is \(x=-1\), and the horizontal asymptote is \(y=-3\). 2. The vertical asymptote shows a shift left by \(1\), replacing \(x\) with \(x+1\). The horizontal asymptote shows a shift down by \(3\). 3. Therefore, \(g(x)=\frac{1}{(x+1)^2}-3\). 4. As a check, \(g(0)=1-3=-2\), which matches the point \((0, -2)\) on the graph.

The vertical asymptote is \(x=-1\), and \(g(x)=\frac{1}{(x+1)^2}-3\).

- Which input makes the denominator zero? - Read both coordinates at the marked point, then substitute them into the function rule. - What happens to the fraction when \(x\) becomes very large in magnitude?

1. The denominator is zero at \(x=-2\), so the domain is \(\mathbb{R}\setminus\{-2\}\). 2. The marked point is \(P(2,-2)\). Substitute its coordinates: \(-2=\frac{a}{2+2}-1\), so \(a=-4\). 3. For very large positive or negative inputs, \(\frac{-4}{x+2}\) approaches \(0\). Therefore, the horizontal asymptote is \(y=-1\).

a) Excluded input: \(x=-2\); domain: \(\mathbb{R}\setminus\{-2\}\) b) \(P(2,-2)\); \(a=-4\) c) \(y=-1\); the fraction approaches \(0\) for large positive or negative inputs.

- The vertical asymptote identifies the excluded input. - Use the horizontal asymptote and a point on the graph to determine the function rule. - Consider what happens to a fraction when its denominator becomes very large.

1. The vertical asymptote is \(x = 4\), so \(x = 4\) is excluded from the domain. Thus, the domain is \(\mathbb{R} \setminus \{4\}\), and \(h = 4\). 2. The horizontal asymptote is \(y = 1\), so \(k = 1\). 3. The point \((6, 0)\) lies on the graph. Substitute it into \(g(x) = \frac{a}{x - 4} + 1\): \(0 = \frac{a}{6 - 4} + 1\). Therefore, \(-1 = \frac{a}{2}\), so \(a = -2\). 4. The function rule is \(g(x) = -\frac{2}{x - 4} + 1\). 5. As \(x \to \infty\), the denominator \(x - 4\) grows without bound, so \(-\frac{2}{x - 4} \to 0\). Therefore, \(g(x) \to 1\). The right branch approaches the horizontal asymptote \(y = 1\) from below.

a) \(\mathbb{R} \setminus \{4\}\) b) \(g(x) = -\frac{2}{x - 4} + 1\) c) As \(x \to \infty\), \(g(x) \to 1\), matching the horizontal asymptote \(y = 1\).

- Determine the upper and lower boundaries of the band. - Evaluate the function at each given x-value. - Compare each result with the band’s interval.

1. The band extends from \(y=1.9\) to \(y=2.1\). 2. \(g(-8)=\frac{-17}{-9}=\frac{17}{9}\approx1.889\). Since \(1.889<1.9\), the point is outside the band. 3. \(g(-10)=\frac{-21}{-11}=\frac{21}{11}\approx1.909\). Since \(1.9<1.909<2.1\), the point is inside the band.

a) \(g(-8)=\frac{17}{9}\approx1.889<1.9\), so the point is outside. b) \(g(-10)=\frac{21}{11}\approx1.909\), which lies between \(1.9\) and \(2.1\), so the point is inside.

- Factor the numerator and denominator, but keep track of which inputs were excluded originally. - Which factor disappears after simplification, and which remains in the denominator? - To find the height of a hole, use the simplified expression at the excluded input. - Compare the leading terms for the horizontal asymptote.

1. Factor the denominator: \(x^2-1=(x-1)(x+1)\). The domain is \(\mathbb{R}\setminus\{-1,1\}\). 2. For allowed inputs, \(h(x)=\frac{2x(x+1)}{(x-1)(x+1)}=\frac{2x}{x-1}\). 3. The canceled factor leaves a hole at \(x=-1\). The simplified expression gives \(\frac{2(-1)}{-1-1}=1\), so the hole is \((-1,1)\). 4. The remaining denominator is zero at \(x=1\), where the numerator is nonzero, giving the vertical asymptote \(x=1\). 5. Both original polynomials have degree \(2\); the leading-coefficient ratio is \(2\), so the horizontal asymptote is \(y=2\).

a) \(\mathbb{R}\setminus\{-1,1\}\) b) Hole: \((-1,1)\); vertical asymptote: \(x=1\) c) \(y=2\)

- What upstream speed remains after subtracting the current from the motorboat's speed? - Think about what happens to travel time when upstream progress becomes very slow. - For part c), use the physical domain when multiplying an inequality by the denominator.

1. The boat must move faster than the current to make progress upstream, so \(x>3\). 2. The denominator approaches zero from the positive side as \(x\) gets closer to \(3\) from above. The time grows without bound. The vertical asymptote is \(x=3\). 3. For \(x>3\), solve \(\frac{18}{x-3}<2\): multiply by the positive denominator to get \(18<2x-6\), or \(x>12\). The least whole-number speed is \(13\,\text{mph}\).

a) \(x>3\). b) \(x=3\); travel time grows without bound as the boat's speed approaches the current's speed from above. c) \(13\,\text{mph}\).

- Replace \(x\) with \(-x\) in the original function. - Find where the shifted denominator equals \(0\). - Check whether excluded domain values occur in opposite pairs.

1. \(h(-x)=\frac{1}{(-x)^2-9}+2=\frac{1}{x^2-9}+2=h(x)\), so \(h\) is even. 2. The shifted function is \(g(x)=\frac{1}{(x-2)^2-9}+2\). 3. Its denominator is \(0\) when \(x-2=\pm3\), so the domain is \(\mathbb{R}\setminus\{-1, 5\}\). 4. The domain of an even function must be symmetric about \(0\). Here, \(5\) is excluded but \(-5\) is included. Therefore, the domain is not symmetric about \(0\), so \(g\) cannot be even.

1) \(h(-x)=h(x)\), so \(h\) is even. 2) The domain of \(g\) is \(\mathbb{R}\setminus\{-1, 5\}\), which is not symmetric about \(0\). Therefore, \(g\) is not even.

- Substitute the given value of \(a\) and inspect the denominator multiplicity. - For which value of \(a\) does the numerator contain the factor \(x-3\)? - After simplifying, how does the exponent on the remaining denominator factor determine whether the graph changes sign?

1. When \(a=0\), \(p_0(x)=\frac{x}{(x-3)^2}\). At \(x=3\), the numerator is \(3\), so no factor cancels. The denominator zero has even multiplicity \(2\). Therefore, \(x=3\) is a vertical asymptote and the graph does not change sign across it. 2. The behavior changes when the numerator also contains the factor \(x-3\), which requires \(a=3\). Then \(p_3(x)=\frac{x-3}{(x-3)^2}=\frac{1}{x-3}\) for \(x\neq 3\). The remaining denominator zero has odd multiplicity \(1\), so the graph changes sign across the vertical asymptote at \(x=3\).

1. For \(a=0\), the vertical asymptote is \(x=3\), and the graph does not change sign across it. 2. The behavior changes when \(a=3\). Then \(x=3\) is still a vertical asymptote, but the graph changes sign across it.

- For which values of \(a\) does \(x^2=a\) have zero, one, or two real solutions? - Can a denominator zero be canceled by the numerator? - Compare the case with one original denominator zero to the case with two original zeros, one canceled.

1. If \(a<0\), the denominator equation \(x^2=a\) has no real solutions. There are no excluded inputs or vertical asymptotes. 2. If \(a=0\), then \(g_0(x)=\frac{x-3}{x^2}\). The denominator has one real zero \(x=0\), and the numerator is nonzero there. Thus \(x=0\) is the only vertical asymptote. 3. If \(a>0\), the denominator has two real zeros \(x=\pm\sqrt{a}\). To have exactly one vertical asymptote, one factor must cancel with the numerator \(x-3\). That requires \(\sqrt{a}=3\), giving \(a=9\). 4. For \(a=9\), the original domain excludes \(x=-3\) and \(x=3\). The expression simplifies to \(\frac{1}{x+3}\) away from \(x=3\), leaving a hole at \(x=3\) and a vertical asymptote at \(x=-3\). 5. The values are \(a=0\) and \(a=9\).

a) If \(a<0\), the denominator \(x^2-a\) is always positive, so there is no vertical asymptote. b) \(a=0\) or \(a=9\). For \(a=0\), the only vertical asymptote is \(x=0\). For \(a=9\), it is \(x=-3\), with a hole at \(x=3\).