Free Math Worksheets

- 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.

- A rational expression is undefined when its denominator is zero. - Solve the denominator equation first. - Check whether the numerator is also zero at the excluded input.

1. The denominator cannot equal zero. Solve \(x + 6 = 0\), which gives \(x = -6\). 2. Therefore, the domain is \(\mathbb{R} \setminus \{-6\}\). 3. At \(x = -6\), the numerator is \(8(-6) - 4 = -52\), which is not zero. Therefore, the nonremovable discontinuity produces a vertical asymptote at \(x = -6\).

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. - The constant outside the fraction gives the horizontal asymptote. - Match each function to the parameters \(h\) and \(k\) in the general form.

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. In general, for \(y = \frac{a}{x - h} + k\), the vertical asymptote is \(x = h\) and the horizontal asymptote is \(y = k\).

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

- 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 the function at \(x = -2\): \(f(-2) = \frac{2}{2} - 3 = -2\). Therefore, \(P\) 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) \(\left(0, -\frac{2}{3}\right)\)

- 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, \(\frac{3}{x + 4} \to 0\), 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.

- 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 \((0, 0)\).

- Find the limit of the fractional term as \(x\to\pm\infty\). - Identify the vertical line about which the graph is symmetric. - Replace \(x\) with \(-x\) and simplify.

1. As \(x\to\pm\infty\), \(\frac{8}{(x+d)^2}\to0\), so the horizontal asymptote is \(y=e\). Therefore, \(e=2\). 2. The vertical asymptote and axis of symmetry are both \(x=-d\). For the axis of symmetry 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? - Factoring out \(x^2\) means dividing each term by \(x^2\). - What happens to \(\frac{1}{x}\) and \(\frac{1}{x^2}\) as \(x\) increases without bound? - A finite limit at infinity gives the height of a horizontal asymptote.

1. Factor \(x^2\) from the numerator: \(10x^2-5x+1=x^2\left(10-\frac{5}{x}+\frac{1}{x^2}\right)\). 2. Factor \(x^2\) from the denominator: \(2x^2+3=x^2\left(2+\frac{3}{x^2}\right)\). 3. For \(x\neq 0\), cancel \(x^2\): \(f(x)=\frac{10-\frac{5}{x}+\frac{1}{x^2}}{2+\frac{3}{x^2}}\). 4. As \(x\to\infty\), the terms containing \(\frac{1}{x}\) or \(\frac{1}{x^2}\) approach \(0\). 5. Therefore, \(\lim_{x\to\infty}f(x)=\frac{10}{2}=5\), and the horizontal asymptote is \(y=5\).

a) \(f(x)=\frac{10-\frac{5}{x}+\frac{1}{x^2}}{2+\frac{3}{x^2}}\) for \(x\neq 0\) b) \(\lim_{x\to\infty}f(x)=5\); horizontal asymptote: \(y=5\)

- What is the end behavior when the denominator has higher degree than the numerator? - Which polynomial grows faster as \(|x|\) becomes large? - Distinguish the unequal-degree case from the equal-degree case.

1. The numerator has degree \(1\), and the denominator has degree \(2\). 2. Since the denominator’s degree is greater, \(\lim_{x\to\pm\infty}\frac{3x+12}{x^2-16}=0\). 3. Therefore, the horizontal asymptote is \(y=0\). 4. The student is correct that a horizontal asymptote exists, but the claimed equation \(y=3\) is incorrect.

The statement is partly incorrect. A horizontal asymptote does exist, but it is \(y=0\), not \(y=3\), because the denominator’s degree is greater than the numerator’s degree.

- Divide the numerator and denominator by the greatest power of \(x\) in the denominator. - What happens to terms such as \(\frac{1}{x}\) and \(\frac{1}{x^2}\) as \(|x|\) becomes large? - Compare the degrees and leading coefficients of the numerator and denominator.

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

a) \(\lim_{x\to\pm\infty}f(x)=-3\); horizontal asymptote: \(y=-3\) b) \(\lim_{x\to\pm\infty}g(x)=0\); horizontal asymptote: \(y=0\) c) \(\lim_{x\to\pm\infty}h(x)=3\); 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 degree of each numerator with the degree of its denominator. - Divide the numerator and denominator by the highest power of \(x\) in the denominator. - Recall what happens to \(\frac{1}{x}\) and \(\frac{1}{x^2}\) as \(x\to\infty\).

1. For \(g\), the numerator and denominator both have degree \(2\). Divide by \(x^2\): \(\frac{4-\frac{2}{x}+\frac{1}{x^2}}{2+\frac{5}{x^2}}\to\frac{4}{2}=2\). 2. For \(h\), the numerator has degree \(1\), while the denominator has degree \(2\). Divide by \(x^2\): \(\frac{\frac{4}{x}-\frac{2}{x^2}}{2+\frac{5}{x^2}}\to\frac{0}{2}=0\). 3. Equal numerator and denominator degrees give the ratio of the leading coefficients. When the denominator degree is greater, the denominator grows faster and the quotient approaches \(0\).

\(\lim_{x\to\infty}g(x)=2\) and \(\lim_{x\to\infty}h(x)=0\). Thus, \(g\) has horizontal asymptote \(y=2\), while \(h\) has horizontal asymptote \(y=0\).

- Factor \(k\) from the denominator. - Use the numerator to find an x-intercept, then check the domain. - Check whether any denominator factor cancels. - Compare the degrees for end behavior.

1. Factor the denominator: \(kx^2-4k=k(x^2-4)=k(x-2)(x+2)\). Since \(k\neq 0\), the denominator is zero only at \(x=-2\) and \(x=2\). Thus, the domain is \(\mathbb{R}\setminus\{-2, 2\}\) for every \(k\). 2. Set the numerator equal to zero: \(2x+6=0\), so \(x=-3\). This value is in the domain, so the x-intercept is \((-3, 0)\). 3. At \(x=-2\) and \(x=2\), the numerator is nonzero, so both values produce vertical asymptotes for every \(k\). 4. The denominator has greater degree than the numerator, so \(\lim_{x\to\pm\infty}g_k(x)=0\). Therefore, the horizontal asymptote is \(y=0\), independent of \(k\).

a) Domain: \(\mathbb{R}\setminus\{-2, 2\}\) b) x-intercept: \((-3, 0)\) c) Vertical asymptotes: \(x=-2\) and \(x=2\); 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: \(y=0\)

- Trace the graph near the far left and far right edges of the coordinate plane. - Identify the fixed y-value that the curve approaches. - A horizontal asymptote has an equation of the form \(y=c\).

1. Following the graph to the right, the y-values approach \(2\). Therefore, \(\lim_{x\to\infty}f(x)=2\). 2. Following the graph to the left, the y-values also approach \(2\). Therefore, \(\lim_{x\to-\infty}f(x)=2\). 3. Since both end limits equal \(2\), the horizontal asymptote is \(y=2\).

a) \(\lim_{x\to\infty}f(x)=2\) and \(\lim_{x\to-\infty}f(x)=2\) 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 \(y = 2\).

- Determine what happens to \(\frac{4}{x+2}\) as the denominator grows. - Analyze the constant and fractional parts separately. - Connect the finite end limit to a horizontal asymptote.

1. As \(x\to\infty\), the denominator \(x+2\) grows without bound, so \(\frac{4}{x+2}\to0\). 2. Therefore, \(\lim_{x\to\infty}g(x)=0.5+0=0.5\). Marie’s claim is false. 3. The horizontal asymptote is the line whose y-value equals the end limit, so its equation is \(y=0.5\).

a) \(\lim_{x\to\infty}g(x)=0.5\), so Marie’s claim is false. b) \(y=0.5\)

- Look for lines that the graph approaches near an excluded x-value or as \(\lvert x\rvert\) increases. - Set the denominator equal to \(0\) to investigate a vertical asymptote. - Compare leading coefficients to find the horizontal asymptote. - The y-intercept occurs when \(x=0\).

1. The graph suggests a vertical asymptote at \(x=-2\) and a horizontal asymptote at \(y=3\). 2. The denominator is zero at \(x=-2\), while the numerator is \(-12\), so \(x=-2\) is a vertical asymptote. The numerator and denominator have the same degree, and the ratio of leading coefficients is \(3\), so the horizontal asymptote is \(y=3\). 3. Evaluate \(g(0)=\frac{-6}{2}=-3\). Therefore, the y-intercept is \((0, -3)\).

1) Vertical asymptote: \(x=-2\); horizontal asymptote: \(y=3\) 2) The denominator is zero at \(x=-2\), and \(\lim_{x\to\pm\infty}g(x)=3\). 3) y-intercept: \((0, -3)\)

- 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 \(x\).

- 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.

- 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.

- 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 \(x = 3\).

\(b = 6\)

- Determine the domain from the original expression. - What happens when \(x = 2\) is substituted into the original denominator? - Canceling a factor does not restore an input that was originally excluded.

1. The domain must be determined from the original expression before canceling factors. 2. The denominator \(x - 2\) equals zero at \(x = 2\), so the original rational expression is undefined there. 3. The simplification \(g(x) = 1\) is valid only for \(x \ne 2\). 4. Therefore, Maya is correct, and 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 \(\mathbb{R} \setminus \{2\}\).

- The constant outside the fraction determines the horizontal asymptote. - Use the given point to determine the remaining parameter. - A rational expression with a nonzero constant numerator cannot equal \(0\).

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 \(y = 4\)

- 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: \((-3, 0)\)

- Substitute the x-value of the vertical asymptote into the denominator. - Set the function equal to the horizontal-asymptote value. - A fraction can equal \(0\) only if its numerator equals \(0\).

1. The vertical asymptote is \(x = 2\). This input is excluded because it makes the denominator zero, so the function has no point with x-coordinate \(2\). 2. To test the horizontal asymptote, set \(f(x) = 1\): \(\frac{3}{x - 2} + 1 = 1\). 3. Subtracting \(1\) gives \(\frac{3}{x - 2} = 0\). 4. A fraction is zero only when its numerator is zero. Since the numerator is the constant \(3\), this equation has no solution. 5. Therefore, the graph intersects neither asymptote.

a) The function is undefined at \(x = 2\), so the graph cannot intersect the vertical asymptote. b) The equation \(\frac{3}{x - 2} + 1 = 1\) reduces to \(\frac{3}{x - 2} = 0\), which has no solution. The graph does not intersect the horizontal asymptote.

- 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 \((4, 0)\).

- 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 \(\left(0, -\frac{20}{3}\right)\).

- A rational expression equals \(0\) only when its numerator equals \(0\). - Find the input that makes each denominator zero.

1. Set \(h_1(x) = 0\): \(0 = \frac{6}{x} - 2\). Then \(2 = \frac{6}{x}\), so \(2x = 6\) and \(x = 3\). The \(x\)-intercept is \((3, 0)\). 2. For \(h_2\), the equation \(0 = \frac{6}{x - 2}\) has no solution. A rational expression equals \(0\) only when its numerator equals \(0\), but the numerator is the nonzero constant \(6\). 3. For \(h_1\), the denominator is zero at \(x = 0\), so \(D_1 = \mathbb{R} \setminus \{0\}\). 4. For \(h_2\), the denominator is zero at \(x = 2\), so \(D_2 = \mathbb{R} \setminus \{2\}\).

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) \(D_1 = \mathbb{R} \setminus \{0\}\) and \(D_2 = \mathbb{R} \setminus \{2\}\).

- 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 \(\left(0, -\frac{3}{2}\right)\).

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

- Substitute the given point to find \(k\). - For the vertical asymptote, consider the denominator together with the condition \(k \neq 0\). - Consider what remains as the fraction approaches \(0\) for large \(|x|\). - Changing only the nonzero numerator parameter does not change the denominator or the vertical shift.

1. Substitute \(P(0, 5)\) into \(h_2\): \(5 = \frac{k}{0 - 1} + 1\). 2. Then \(5 = -k + 1\), so \(k = -4\). 3. For \(h_1\), the numerator is the nonzero constant \(4\). For \(h_2\), the condition \(k \neq 0\) also keeps the numerator nonzero. As \(x\) approaches \(1\), the denominator \(x - 1\) approaches \(0\), so the rational term becomes unbounded. Therefore, both graphs have the vertical asymptote \(x = 1\). 4. As \(|x|\) becomes large, each fraction approaches \(0\), leaving the constant \(1\). Therefore, both graphs have the horizontal asymptote \(y = 1\). 5. Changing a nonzero value of \(k\) changes the scale and possibly the orientation of the branches, but not the asymptotes.

a) \(k = -4\) b) Both graphs have the vertical asymptote \(x = 1\) and the horizontal asymptote \(y = 1\). The numerator parameter does not change their locations.

- 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 \((0, -1)\).

- 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 \(f(x)=\frac{3x^2+1}{x^2+1}\).

- What happens to a constant divided by a power of \(x\) as \(|x|\) grows? - When factoring \(x^3\) from the denominator, what remains from the \(x^2\) term? - How is a horizontal asymptote defined using a limit? - Check the result by comparing the leading coefficients.

1. Factor \(x^3\) from the numerator: \(4-2x^3=x^3\left(\frac{4}{x^3}-2\right)\). 2. Factor \(x^3\) from the denominator: \(0.5x^3+x^2-1=x^3\left(0.5+\frac{1}{x}-\frac{1}{x^3}\right)\). 3. For \(x\neq 0\), cancel \(x^3\): \(g(x)=\frac{\frac{4}{x^3}-2}{0.5+\frac{1}{x}-\frac{1}{x^3}}\). 4. As \(x\to-\infty\), \(\frac{4}{x^3}\to 0\), \(\frac{1}{x}\to 0\), and \(\frac{1}{x^3}\to 0\). 5. Therefore, \(\lim_{x\to-\infty}g(x)=\frac{-2}{0.5}=-4\), 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 \(x\neq 0\) b) \(\lim_{x\to-\infty}g(x)=-4\); horizontal asymptote: \(y=-4\)

- How does the value of \(a\) affect the degree and leading coefficient of the numerator? - What determines the end behavior when the numerator and denominator have the same degree? - What happens when the denominator has greater degree than the numerator?

1. When \(a\neq 0\), the numerator and denominator both have degree \(2\), so the limit is the ratio of the leading coefficients: \(\lim_{x\to\infty}f_a(x)=\frac{a}{3}\). 2. Set \(\frac{a}{3}=4\). Solving gives \(a=12\). 3. When \(a=0\), \(f_0(x)=\frac{6x}{3x^2-12}\). The denominator has greater degree than the numerator. 4. Therefore, \(\lim_{x\to\pm\infty}f_0(x)=0\), so the horizontal asymptote is \(y=0\).

a) \(a=12\) b) \(f_0(x)\to 0\) as \(x\to\pm\infty\); the horizontal asymptote is \(y=0\).

- Factor the numerator and denominator before identifying asymptotes. - Decide whether each excluded value produces a removable discontinuity or a vertical asymptote. - Examine the function as \(x\to\pm\infty\).

1. Factor the numerator and denominator: \(g(x)=\frac{3(x-2)}{(x-2)(x+2)}\). 2. For \(x\neq 2\), this simplifies to \(g(x)=\frac{3}{x+2}\). 3. The factor \(x-2\) cancels, so \(x=2\) is a removable discontinuity, not a vertical asymptote. The remaining denominator is zero at \(x=-2\), so the vertical asymptote is \(x=-2\). 4. Since \(\frac{3}{x+2}\to 0\) as \(x\to\pm\infty\), 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. - To find when the cost is below a threshold, write and solve an inequality. - Remember that the number of servers must be a whole number. - For the limit, compare the leading coefficients. - Interpret the asymptote using the units and context of the model.

1. Evaluate the function: \(B(20)=\frac{6000}{17}\approx\$352.94\), \(B(100)=\frac{20400}{65}\approx\$313.85\), and \(B(500)=\frac{92400}{305}\approx\$302.95\). The average cost per server decreases as the number of servers increases. 2. Solve \(\frac{180x+2400}{0.6x+5}<320\). Since \(0.6x+5>0\) for \(x\ge 10\), this is equivalent to \(180x+2400<192x+1600\), so \(800<12x\), or \(x>\frac{200}{3}\). 3. Because \(x\) must be a whole number, the least possible value is \(67\). 4. The numerator and denominator have the same degree, so \(\lim_{x\to\infty}B(x)=\frac{180}{0.6}=300\). The horizontal asymptote is \(y=300\). In the model, the average monthly cost per server approaches \(\$300\) as the number of servers grows. Because \(B(x)-300=\frac{900}{0.6x+5}>0\), the modeled cost remains above \(\$300\).

a) \(B(20)\approx\$352.94\), \(B(100)\approx\$313.85\), and \(B(500)\approx\$302.95\). The cost per server decreases as the number of servers increases. b) \(67\) servers c) \(\lim_{x\to\infty}B(x)=300\). The horizontal asymptote is \(y=300\), meaning the modeled average cost approaches \(\$300\) per server from above.

- Pay close attention to the unit in which \(N(t)\) is measured. - Substitute \(t=0\) to find the initial value. - Convert the target population to hundreds before solving an equation. - For the limit, compare the leading coefficients.

1. Evaluate \(N(0)=\frac{50}{5}=10\). Since \(N\) is measured in hundreds, the initial population is \(10\cdot 100=1000\) animals. 2. A population of \(5500\) animals corresponds to \(N(t)=55\). Solve \(\frac{120t+50}{2t+5}=55\). 3. This gives \(120t+50=110t+275\), so \(10t=225\) and \(t=22.5\). 4. The numerator and denominator have the same degree, so \(\lim_{t\to\infty}N(t)=\frac{120}{2}=60\). The horizontal asymptote is \(y=60\), representing a long-term population of \(6000\) animals.

a) \(1000\) animals b) \(22.5\) years c) The horizontal asymptote is \(y=60\). The modeled population approaches a carrying capacity of \(6000\) animals.

- Begin by finding the zeros of each denominator. - Check whether the numerator is also zero at any excluded value. - After simplifying, note the multiplicity of each zero that remains in the denominator. - How does an even or odd denominator multiplicity affect the sign of the function across a vertical asymptote?

1. For part a, the denominator has a zero at \(x=1\) with multiplicity \(2\). The numerator is \(5\) at \(x=1\), so \(x=1\) is a vertical asymptote. Because the multiplicity is even, the graph does not change sign across it. 2. For part b, factor the denominator: \(x^2-4x=x(x-4)\). The zeros \(x=0\) and \(x=4\) each have multiplicity \(1\), and the numerator is never zero. Both values are vertical asymptotes, and the graph changes sign across each one because each multiplicity is odd. 3. For part c, factor the denominator: \(x^2-4=(x-2)(x+2)\). Canceling the common factor \(x-2\) gives \(h(x)=\frac{1}{x+2}\) for \(x\neq 2\). Thus, \(x=2\) is a removable discontinuity, not a vertical asymptote. The value \(x=-2\) is a vertical asymptote with odd multiplicity \(1\), so the graph changes sign across it.

a) \(x=1\) is a vertical asymptote; the graph does not change sign across it. b) \(x=0\) and \(x=4\) are vertical asymptotes; the graph changes sign across both. c) \(x=-2\) is a vertical asymptote; the graph changes sign across it. The value \(x=2\) is a removable discontinuity.

- Factor the numerator and denominator before classifying the discontinuities. - What happens when the same factor makes both the numerator and denominator zero? - After simplifying, what is the multiplicity of the factor that remains in the denominator?

1. Factor the denominator: \(x^2-x-6=(x-3)(x+2)\). Therefore, the excluded values are \(x=3\) and \(x=-2\). 2. Factor the numerator: \(2x-6=2(x-3)\). 3. Cancel the common factor for \(x\neq 3\): \(f(x)=\frac{2(x-3)}{(x-3)(x+2)}=\frac{2}{x+2}\). Because the factor causing the exclusion at \(x=3\) cancels, \(x=3\) is a removable discontinuity. 4. The factor \(x+2\) remains in the denominator with multiplicity \(1\), so \(x=-2\) is a vertical asymptote. Its multiplicity is odd, so the graph changes sign across the asymptote.

The discontinuities occur at \(x=3\) and \(x=-2\). \(x=3\) is a removable discontinuity. \(x=-2\) is a vertical asymptote, and the graph changes sign across it.

- 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 and denominator completely. - Can the rational expression be simplified? - After canceling all common factors, examine the exponent on the factor that remains in the denominator. - What does an even or odd denominator multiplicity tell you about the sign of the graph near a vertical asymptote?

1. Factor the numerator using the difference of squares: \(x^2-9=(x-3)(x+3)\). 2. The denominator is zero at \(x=3\), so this value is excluded from the domain. 3. Simplify for \(x\neq 3\): \(k(x)=\frac{(x-3)(x+3)}{(x-3)^3}=\frac{x+3}{(x-3)^2}\). 4. The remaining denominator factor has multiplicity \(2\), which is even. 5. Therefore, \(x=3\) is a vertical asymptote and the graph does not change sign across it.

\(x=3\) is a vertical asymptote where the graph does not change sign. After simplifying, \(k(x)=\frac{x+3}{(x-3)^2}\) for \(x\neq 3\). The remaining denominator zero has even multiplicity \(2\).

- First find the real zeros of each denominator. - Check whether the numerator is also zero at any excluded value, and simplify when possible. - Use the multiplicity of each denominator zero after simplification. - What can you conclude when the denominator is never zero for real \(x\)?

1. For part a, factor: \(3x-12=3(x-4)\) and \(x^2-16=(x-4)(x+4)\). For \(x\neq 4\), \(f(x)=\frac{3}{x+4}\). Therefore, \(x=4\) is a removable discontinuity. The value \(x=-4\) is a vertical asymptote with odd multiplicity \(1\), so the graph changes sign across it. 2. For part b, \(x^2+4x+4=(x+2)^2\). For \(x\neq -2\), \(g(x)=\frac{x+2}{(x+2)^2}=\frac{1}{x+2}\). Thus, \(x=-2\) is a vertical asymptote with odd multiplicity \(1\), and the graph changes sign across it. 3. For part c, \(x^2+1>0\) for every real \(x\). The function has no real discontinuities and no vertical asymptotes.

a) \(x=-4\) is a vertical asymptote, and the graph changes sign across it. The value \(x=4\) is a removable discontinuity. b) \(x=-2\) is a vertical asymptote, and the graph changes sign across it. c) There are no vertical asymptotes or removable discontinuities.

- 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 \(k>0\).

- Determine the excluded inputs first. - Set \(g(x)=3\) and clear the denominator carefully. - Rewrite \(g(x)\) to make its horizontal asymptote visible. - Analyze the sign of the denominator on the intervals separated by the vertical asymptotes.

1. The domain is \(\mathbb{R}\setminus\{-1, 1\}\). 2. To test for an intersection with \(y=3\), solve \(\frac{3x^2}{x^2-1}=3\). Multiplying by \(x^2-1\), which is nonzero on the domain, gives \(3x^2=3x^2-3\), or \(0=-3\). Therefore, there is no intersection. 3. For \(|x|>1\), rewrite the function as \(g(x)=3+\frac{3}{x^2-1}\). Since \(x^2-1>0\), \(g(x)>3\). The values approach \(3\) but never equal it, and increase without bound near \(x=\pm1\), so this part of the range is \((3, \infty)\). 4. For \(-1<x<1\), the denominator is negative and the numerator is nonnegative, so \(g(x)\le0\). The value \(0\) occurs at \(x=0\), and the function decreases without bound near the vertical asymptotes. Thus, this part of the range is \((-\infty, 0]\). 5. Combining the intervals, the range is \((-\infty, 0]\cup(3, \infty)\).

a) The equation \(g(x)=3\) leads to the contradiction \(0=-3\), so there is no intersection. b) \((-\infty, 0]\cup(3, \infty)\)

- First identify the values that make each denominator zero. - Factor the expressions and cancel common factors when possible. - After simplifying, is the exponent on the remaining denominator factor even or odd? - What would it mean if no factor remained in the denominator after cancellation?

1. For part a, the excluded value is \(x=2\). Factor and simplify: \(f(x)=\frac{2(x-2)}{(x-2)^3}=\frac{2}{(x-2)^2}\) for \(x\neq 2\). The remaining denominator zero has even multiplicity \(2\), so \(x=2\) is a vertical asymptote where the graph does not change sign. 2. For part b, the excluded value is \(x=-1\). Factor and simplify: \(g(x)=\frac{(x-1)(x+1)}{(x+1)^2}=\frac{x-1}{x+1}\) for \(x\neq -1\). The remaining denominator zero has odd multiplicity \(1\), so \(x=-1\) is a vertical asymptote where the graph changes sign.

a) \(x=2\) is a vertical asymptote where the graph does not change sign. b) \(x=-1\) is a vertical asymptote where the graph changes sign.

- Find the zeros of the denominator to determine the domain. - Check whether any denominator zero is also a numerator zero. - What does complete cancellation of a denominator factor imply about the discontinuity? - The equation of a vertical asymptote uses the \(x\)-value where an uncanceled denominator factor is zero.

1. Factor the denominator: \(x^2-9=(x-3)(x+3)\). Therefore, \(D_h=\mathbb{R}\setminus\{-3, 3\}\). 2. Factor the numerator: \(x^2-2x-3=(x-3)(x+1)\). 3. For \(x\neq 3\), simplify: \(h(x)=\frac{(x-3)(x+1)}{(x-3)(x+3)}=\frac{x+1}{x+3}\). Because the factor \(x-3\) cancels completely, \(x=3\) is a removable discontinuity, not a vertical asymptote. 4. The factor \(x+3\) remains in the denominator with odd multiplicity \(1\). Thus, \(x=-3\) is a vertical asymptote and the graph changes sign across it.

\(D_h=\mathbb{R}\setminus\{-3, 3\}\) The only vertical asymptote is \(x=-3\), and the graph changes sign across it. The value \(x=3\) is a removable discontinuity.

- Compare the degrees of the numerator and denominator. - When the denominator has greater degree, what is the limit? - When the degrees are equal, compare the leading coefficients.

1. For \(f_2\) and \(f_4\), the numerator has lower degree than the denominator, so \(\lim_{x\to\pm\infty}f_2(x)=0\) and \(\lim_{x\to\pm\infty}f_4(x)=0\). 2. For \(f_1\), \(f_3\), and \(f_5\), the numerator and denominator have the same degree. Their limits are the ratios of leading coefficients. 3. Thus, \(\lim_{x\to\pm\infty}f_1(x)=\frac{3}{1}=3\), \(\lim_{x\to\pm\infty}f_3(x)=\frac{6}{2}=3\), and \(\lim_{x\to\pm\infty}f_5(x)=\frac{-9}{-3}=3\). 4. Therefore, \(f_2\) and \(f_4\) have horizontal asymptote \(y=0\), while \(f_1\), \(f_3\), and \(f_5\) have horizontal asymptote \(y=3\).

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

- First find the zeros of each denominator. - Check whether each denominator zero is also a numerator zero. - Factor the numerator and denominator completely, then cancel common factors. - What does it mean if a denominator factor cancels completely or remains after simplification?

1. For part a, factor both expressions: \(x^2-9=(x-3)(x+3)\) and \(x^2-2x-3=(x-3)(x+1)\). The excluded values are \(x=3\) and \(x=-1\). For \(x\neq 3\), \(f(x)=\frac{x+3}{x+1}\). Thus, \(x=3\) is a removable discontinuity, while \(x=-1\) is an infinite discontinuity. 2. For part b, factor both expressions: \(x^3-4x=x(x-2)(x+2)\) and \(x^2-2x=x(x-2)\). The excluded values are \(x=0\) and \(x=2\). For \(x\neq 0,2\), \(g(x)=x+2\). Both denominator factors cancel completely, so both discontinuities are removable.

a) \(x=3\) is a removable discontinuity, and \(x=-1\) is an infinite discontinuity. b) \(x=0\) and \(x=2\) are both removable discontinuities.

- How do you determine the domain of a rational function? - Use factoring identities to rewrite each numerator and denominator. - What happens when a denominator zero has greater multiplicity than the same zero in the numerator? - When can a discontinuity be removed by canceling a common factor?

1. For part a, factor: \(x^2-2x=x(x-2)\) and \(x^2-4x+4=(x-2)^2\). The domain is \(D_h=\mathbb{R}\setminus\{2\}\). For \(x\neq 2\), \(h(x)=\frac{x}{x-2}\). One factor of \(x-2\) remains in the denominator, so \(x=2\) is an infinite discontinuity and the graph changes sign across its vertical asymptote. 2. For part b, factor: \(x^2+x-6=(x+3)(x-2)\) and \(x^2+3x=x(x+3)\). The domain is \(D_k=\mathbb{R}\setminus\{-3, 0\}\). For \(x\neq -3\), \(k(x)=\frac{x-2}{x}\). Therefore, \(x=-3\) is a removable discontinuity, while \(x=0\) is an infinite discontinuity and the graph changes sign across its vertical asymptote.

a) \(D_h=\mathbb{R}\setminus\{2\}\). The discontinuity at \(x=2\) is infinite, and the graph changes sign across the vertical asymptote. b) \(D_k=\mathbb{R}\setminus\{-3, 0\}\). The discontinuity at \(x=-3\) is removable. The discontinuity at \(x=0\) is infinite, and the graph changes sign across the vertical asymptote.

- What happens when both the numerator and denominator are zero at the same input? - Factor each expression completely to see whether a denominator factor cancels. - Check each excluded value separately. - Compare the multiplicity of each common zero in the numerator and denominator.

1. For \(f\), factor the denominator as \(x(x-3)\). Factor the numerator as \(x(x^2-2x-3)=x(x-3)(x+1)\). Both denominator factors cancel completely, so the removable discontinuities are \(x=0\) and \(x=3\). 2. For \(g\), factor the denominator as \((x-2)(x+2)\). Factor the numerator as \(0.2(x^2-4x+4)=0.2(x-2)^2\). One factor of \(x-2\) cancels completely from the denominator, so \(x=2\) is removable. The factor \(x+2\) does not cancel, so \(x=-2\) is not removable.

The removable discontinuities of \(f\) are \(x=0\) and \(x=3\). The removable discontinuity of \(g\) is \(x=2\).

- Rewrite the numerator using the denominator. - Determine the possible values of \(x^2+2\). - Track which endpoint values are reached and which are only approached.

1. Rewrite the function: \(g(x)=\frac{x^2+2-6}{x^2+2}=1-\frac{6}{x^2+2}\). 2. Since \(x^2\ge0\), the denominator \(x^2+2\) takes all values in \([2, \infty)\). 3. Therefore, \(\frac{6}{x^2+2}\) takes all values in \((0, 3]\). The value \(3\) occurs at \(x=0\), while \(0\) is approached but never reached. 4. Subtracting these values from \(1\) gives \(-2\le g(x)<1\). Hence, the range is \([-2, 1)\).

\([-2, 1)\)

- 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) \((1, 1)\)

- Determine when the denominator can equal zero. - Replace \(x\) with \(-x\) to test symmetry. - Set the numerator equal to zero to find x-intercepts. - Compare the leading terms or divide by \(x^2\) to find the horizontal asymptote.

1. The denominator cannot be zero. If \(a>0\), then \(x^2+a>0\) for every real \(x\), so the domain is \(\mathbb R\). If \(a<0\), the denominator is zero at \(x=\pm\sqrt{-a}\). Therefore, the statement depends on \(a\) and is true exactly when \(a>0\). 2. Since \(g_a(-x)=g_a(x)\), every graph is symmetric about the y-axis. The statement is true for every allowed \(a\). 3. The numerator is zero when \(x^2=a\). There are two real solutions when \(a>0\) and none when \(a<0\). Therefore, the statement depends on \(a\) and is true exactly when \(a>0\). 4. Divide numerator and denominator by \(x^2\): \(g_a(x)=\frac{1-a/x^2}{1+a/x^2}\). As \(x\to\pm\infty\), this expression approaches \(1\). Thus, \(y=1\) is a horizontal asymptote for every allowed \(a\).

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 the numerator and denominator completely. - A canceled denominator factor creates a removable discontinuity; an uncanceled denominator factor creates a vertical asymptote. - Check each numerator zero against the original domain. - Use the simplified expression to find the y-coordinate of a hole.

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. For domain values, the function simplifies to \(\frac{x+1}{x+2}\). The uncanceled denominator factor gives a vertical asymptote at \(x=-2\). 3. The canceled factor gives a removable discontinuity at \(x=2\). The missing point has y-coordinate \(\frac{2+1}{2+2}=\frac{3}{4}\), so the hole is at \(\left(2, \frac{3}{4}\right)\). 4. The original numerator is zero at \(x=2\) and \(x=-1\), but \(x=2\) is not in the domain. Therefore, the only x-intercept is \((-1, 0)\).

Domain: \(\mathbb{R}\setminus\{-2, 2\}\); vertical asymptote: \(x=-2\); removable discontinuity at \(\left(2, \frac{3}{4}\right)\); 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 \(a=3\).

- Complete the square in the denominator. - Compare the function values at \(1+h\) and \(1-h\). - Use the fact that \((-h)^2=h^2\).

1. Complete the square in the denominator: \(x^2-2x+5=(x-1)^2+4\). Thus, \(f(x)=\frac{8}{(x-1)^2+4}\). 2. To verify the symmetry algebraically, compare inputs equally far from \(1\): \(f(1+h)=\frac{8}{h^2+4}\) and \(f(1-h)=\frac{8}{(-h)^2+4}=\frac{8}{h^2+4}\). 3. Since \(f(1+h)=f(1-h)\) for every real \(h\), the graph is symmetric about \(x=1\).

\(f(1+h)=f(1-h)=\frac{8}{h^2+4}\) for every real \(h\), so the graph is symmetric about \(x=1\).

- 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 \(x>20.10\)

- Set the numerator equal to zero and consider the sign of \(a\). - Check whether each possible numerator zero belongs to the domain. - Determine when the numerator and denominator vanish at the same values. - A canceled factor produces a removable discontinuity.

1. The numerator is zero when \(x^2=a\). 2. If \(a<0\), there are no real numerator zeros. If \(a=0\), \(x=0\) is a double zero. If \(a>0\), the possible zeros are \(x=\pm\sqrt{a}\). 3. The denominator is zero at \(x=\pm 3\), so those values are excluded from every domain. When \(a=9\), the possible numerator zeros coincide with the excluded values and are not x-intercepts. In fact, \(f_9(x)=-1\) for \(x\neq\pm 3\). 4. Therefore, if \(a>0\) and \(a\neq 9\), the x-intercepts are \((-\sqrt{a}, 0)\) and \((\sqrt{a}, 0)\). If \(a<0\) or \(a=9\), there are no x-intercepts; if \(a=0\), the only x-intercept is \((0, 0)\), with multiplicity \(2\). 5. When \(a\neq 9\), the numerator is nonzero at \(x=\pm 3\), so both discontinuities are vertical asymptotes with sign changes. When \(a=9\), the common factors cancel, so \(x=-3\) and \(x=3\) are removable discontinuities with missing y-value \(-1\).

X-intercepts: - If \(a<0\) or \(a=9\): none - If \(a=0\): \((0, 0)\), with multiplicity \(2\) - If \(a>0\) and \(a\neq 9\): \((-\sqrt{a}, 0)\) and \((\sqrt{a}, 0)\) Discontinuities at \(x=\pm 3\): vertical asymptotes with sign changes when \(a\neq 9\), and removable discontinuities when \(a=9\)

- 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 \(g(x)=\frac{1}{(x+1)^2}-3\).

- Find the value of \(x\) that makes the denominator zero. - A point on the graph must satisfy the function equation. Substitute both coordinates of \(P\). - For large values of \(|x|\), determine what happens to the fraction in the function rule.

1. The denominator is zero when \(x + 2 = 0\), so the excluded input is \(x = -2\). Thus, the domain is \(D = \mathbb{R} \setminus \{-2\}\). 2. Because \(P(2, -2)\) lies on the graph, substitute \(x = 2\) and \(g(x) = -2\): \(-2 = \frac{a}{2 + 2} - 1\). Then \(-1 = \frac{a}{4}\), so \(a = -4\). 3. As \(|x|\) becomes large, \(\frac{-4}{x + 2}\) approaches \(0\). Therefore, \(g(x)\) approaches \(-1\), so the horizontal asymptote is \(y = -1\).

a) Excluded input: \(x = -2\); domain: \(D = \mathbb{R} \setminus \{-2\}\). b) \(a = -4\). c) The horizontal asymptote is \(y = -1\) because the rational term approaches \(0\) as \(|x|\) becomes large.

- 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 \(y = 1\).

- 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 \((-1, 0)\).

- 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.

- Look for the x-coordinate of the graph’s highest point. - Substitute inputs the same distance to the left and right of that x-value. - Expand and simplify each denominator.

1. The graph reaches its maximum at \(x=3\) and appears mirrored across that vertical line, so \(x_0=3\). 2. Evaluate the two symmetric inputs: \(h(3-a)=\frac{2}{(3-a)^2-6(3-a)+10}=\frac{2}{a^2+1}\). 3. Similarly, \(h(3+a)=\frac{2}{(3+a)^2-6(3+a)+10}=\frac{2}{a^2+1}\). 4. Therefore, \(h(3-a)=h(3+a)\) for every real \(a\), proving symmetry about \(x=3\).

a) \(x=3\) b) Both \(h(3-a)\) and \(h(3+a)\) simplify to \(\frac{2}{a^2+1}\), so the graph is symmetric about \(x=3\).

- Set the denominator equal to zero to find the excluded values. - Factor the difference of squares in the denominator. - A denominator factor that cancels completely creates a removable discontinuity. - Compare the degrees and leading coefficients to find the horizontal asymptote.

1. The denominator factors as \(x^2-1=(x-1)(x+1)\), so the domain is \(D_h=\mathbb{R}\setminus\{-1, 1\}\). 2. Factor and simplify for \(x\neq -1\): \(h(x)=\frac{2x(x+1)}{(x-1)(x+1)}=\frac{2x}{x-1}\). 3. Since the factor \(x+1\) cancels, \(x=-1\) is a removable discontinuity. The corresponding \(y\)-value from the simplified expression is \(\frac{2(-1)}{-1-1}=1\), so the open point is \((-1, 1)\). 4. The factor \(x-1\) remains in the denominator, so \(x=1\) is a vertical asymptote. 5. The numerator and denominator of the original function have the same degree. The ratio of leading coefficients is \(2\), so the horizontal asymptote is \(y=2\).

a) \(D_h=\mathbb{R}\setminus\{-1, 1\}\) b) The removable discontinuity is \((-1, 1)\). The vertical asymptote is \(x=1\). c) The horizontal asymptote is \(y=2\).

- Set the function equal to an arbitrary output \(y\). - Solve the equation for \(x\). - Identify the output that makes the resulting expression undefined. - Verify that the calculated input is never the excluded domain value.

1. Let \(y=\frac{3x}{x+4}\) and solve for \(x\): \(y(x+4)=3x\). 2. Rearranging gives \(x(y-3)=-4y\), so \(x=\frac{-4y}{y-3}\). 3. This expression is defined for every real \(y\) except \(y=3\). Indeed, substituting \(y=3\) into the original equation gives \(3x+12=3x\), a contradiction. 4. For \(y\ne3\), the resulting input can never equal the excluded value \(-4\). Otherwise, \(-4=\frac{-4y}{y-3}\) would imply \(12=0\). 5. Therefore, every real value except \(3\) occurs, and the range is \(\mathbb{R}\setminus\{3\}\).

\(\mathbb{R}\setminus\{3\}\)

- 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 \(x - b \neq 0\), which means \(x \neq b\). 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 \(b \neq 0\), 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.

- 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 \(x=3\). The student’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 \(x-3\).

- 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.

- When does a quadratic equation have zero, one, or two real solutions? - What distinguishes a vertical asymptote from a removable discontinuity? - What happens when a denominator zero is also a numerator zero? - Do not overlook the case in which the denominator has only one distinct real zero.

1. If \(a<0\), the equation \(x^2-a=0\) is equivalent to \(x^2=a\), which has no real solutions because \(a\) is negative. The denominator is never zero, so there are no 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 \(-3\) there. Thus, \(x=0\) is the only vertical asymptote. 3. If \(a>0\), the denominator factors as \((x-\sqrt{a})(x+\sqrt{a})\), so it has two real zeros. To leave exactly one vertical asymptote, one denominator zero must also be the numerator zero \(x=3\), creating a removable discontinuity. 4. The condition \(\sqrt{a}=3\) gives \(a=9\). For this value, \(g_9(x)=\frac{x-3}{(x-3)(x+3)}=\frac{1}{x+3}\) for \(x\neq 3\). There is a removable discontinuity at \(x=3\) and one vertical asymptote at \(x=-3\). 5. Therefore, the required parameter values are \(a=0\) and \(a=9\).

a) For \(a<0\), the equation \(x^2=a\) has no real solutions, so the denominator is never zero and the graph has no vertical asymptotes. b) \(a=0\) and \(a=9\). For \(a=0\), the only vertical asymptote is \(x=0\). For \(a=9\), \(x=-3\) is a vertical asymptote and \(x=3\) is a removable discontinuity.

- Establish a lower bound using the signs of the numerator and denominator. - Rewrite the function as a constant minus a positive fraction. - To prove every value in the interval occurs, set the function equal to an arbitrary \(y\) in that interval and solve for \(x^2\).

1. Since \(x^2\ge0\) and \(x^2+1>0\), \(k(x)\ge0\). The value \(0\) occurs at \(x=0\). 2. Rewrite the function as \(k(x)=2-\frac{2}{x^2+1}\). The fraction is always positive, so \(k(x)<2\). As \(|x|\to\infty\), the fraction approaches \(0\), so the outputs approach \(2\) without reaching it. 3. To show that every value between the bounds occurs, let \(0\le y<2\). Solving \(y=\frac{2x^2}{x^2+1}\) gives \(x^2=\frac{y}{2-y}\). 4. Because \(\frac{y}{2-y}\ge0\), a real solution exists, for example \(x=\sqrt{\frac{y}{2-y}}\). Thus, every \(y\in[0, 2)\) is attained.

\([0, 2)\)