- Compare the solid curve with each dashed line at the far left and far right. - Which line gets progressively closer to the curve?

1. Far to the left and right, the solid curve gets closer to \(l_1\). 2. The distance between the curve and \(l_2\) does not shrink to zero at either end. 3. Therefore \(l_1\) is the slant asymptote.

\(l_1\); the distance between the curve and this line shrinks toward zero at both ends.

- Compare the numerator degree with the denominator degree in each choice. - A linear quotient from polynomial division requires a particular degree difference.

1. A linear slant asymptote occurs when the numerator degree is exactly one greater than the denominator degree. 2. Only choice a) has degree difference \(2-1=1\).

a) \(\frac{x^2+1}{x+1}\)

- Look for a term that is already a linear expression in \(x\). - What happens to the size of a fraction with a fixed numerator and a growing denominator? - Decide which part remains important far from the excluded input.

1. The polynomial part is \(2x-1\), which gives the line \(y=2x-1\). 2. The leftover fraction is \(\frac{5}{x-4}\). As the size of \(x\) grows, the denominator becomes large and the fraction becomes small. 3. Therefore, \(y=2x-1\) is the slant asymptote.

Slant asymptote: \(y=2x-1\). Leftover fraction: \(\frac{5}{x-4}\).

- Find the highest power in each numerator and denominator. - Compare how many degrees apart those powers are. - A slant asymptote must be a nonhorizontal straight line rather than a constant or a curve.

1. In \(f\), the numerator degree is \(2\) and the denominator degree is \(1\). The quotient has degree \(1\), so the asymptote is slant. 2. In \(g\), the degrees are equal, so the quotient is constant and the asymptote is horizontal. 3. In \(h\), the degree difference is \(2\), so division gives a quadratic polynomial part, not a slant line.

a) \(f\) only.

- Do not decide just because a line is dashed; compare its position with the curve. - Look at the gap far to the left and far to the right in each panel. - What must happen to the vertical gap for a line to be an asymptote?

1. In panel a), the dashed line is \(y=x\). The curve follows \(y=x+\frac{2}{x-1}\), so its vertical difference from that line is \(\frac{2}{x-1}\), which becomes small at both distant ends. 2. In panel b), the dashed line is \(y=x+2\). The difference is \(\frac{2}{x-1}-2\), which approaches \(-2\), not zero, at both distant ends. 3. Only panel a) shows the slant asymptote.

Panel a). Its vertical gap shrinks toward \(0\) at both ends. In panel b), the gap approaches \(2\) units instead.

- Compare the curve's height with the dashed line on each far end. - Keep the left and right branches separate when describing their positions. - Above or below is decided by the sign of the vertical difference.

1. The dashed line is \(y=x\), and the curve is \(y=x-\frac{2}{x+1}\). 2. For \(x<-1\), the fraction \(-\frac{2}{x+1}\) is positive, so the curve is above the line. 3. For \(x>-1\), the difference is negative, so the curve is below the line. Far away on both sides, the difference becomes small.

From above on the left; from below on the right.

- Compare the direction of the dashed line in each panel. - Read the height of the level line from the vertical axis. - Use the tilt and intercept of the other line to write its equation.

1. In panel a), the dashed line stays at height \(1\), so the horizontal asymptote is \(y=1\). 2. In panel b), the dashed line rises one unit for each unit to the right and passes through the origin. The slant asymptote is \(y=x\).

a) Horizontal asymptote: \(y=1\). b) Slant asymptote: \(y=x\).

- Which part of the function approaches \(0\) as \(|x|\) becomes large? - How does a line with positive slope behave as \(x\to\infty\) and as \(x\to-\infty\)? - Check the sign of the fractional term for large positive and negative values of \(x\).

1. Since \(\frac{2}{x+4}\to 0\) as \(x\to\pm\infty\), the slant asymptote is \(y=3x+1\). 2. As \(x\to\infty\), \(f(x)\to\infty\). Because \(\frac{2}{x+4}>0\) for large positive \(x\), \(f(x)<3x+1\), so the graph approaches the asymptote from below. 3. As \(x\to-\infty\), \(f(x)\to-\infty\). Because \(\frac{2}{x+4}<0\) for large negative \(x\), \(f(x)>3x+1\), so the graph approaches the asymptote from above.

The slant asymptote is \(y=3x+1\). As \(x\to\infty\), \(f(x)\to\infty\), and the graph approaches the asymptote from below. As \(x\to-\infty\), \(f(x)\to-\infty\), and the graph approaches the asymptote from above.

- What do you multiply the divisor by to obtain the leading term of the numerator? - Subtract that product to find the leftover amount. - Which part of the result remains a straight line when the fraction becomes small?

1. The division has quotient \(x\) because \(x(x+2)=x^2+2x\). 2. Subtracting from the numerator leaves remainder \(3\), so \(f(x)=x+\frac{3}{x+2}\). 3. The remainder fraction becomes small for inputs far from \(-2\), so the slant asymptote is \(y=x\).

\(f(x)=x+\frac{3}{x+2}\); slant asymptote \(y=x\).

- Start with the highest-degree terms in the numerator and denominator. - Subtract the product of each quotient term and the divisor, keeping track of signs. - The slant asymptote comes from the polynomial quotient, not the remainder.

1. Dividing the leading terms gives \(2x\). Subtract \(2x(x-1)=2x^2-2x\) to leave \(3x+3\). 2. Divide \(3x\) by \(x\) to get \(3\). Subtract \(3(x-1)=3x-3\) to leave remainder \(6\). 3. Thus, \(f(x)=2x+3+\frac{6}{x-1}\). The slant asymptote is \(y=2x+3\).

\(f(x)=2x+3+\frac{6}{x-1}\); slant asymptote \(y=2x+3\).

- Which part of the expression describes the straight line approached at distant inputs? - When would the remaining fraction be zero? - Check whether the fraction changes sign at the meeting point.

1. Since \(\frac{x-2}{x^2+1}\) becomes small for large positive or negative \(x\), the slant asymptote is \(y=x+1\). 2. The graph meets this line when \(\frac{x-2}{x^2+1}=0\). Since the denominator is always positive, the equality holds at \(x=2\). 3. At this input the line has height \(3\). The numerator \(x-2\) changes sign there, so the graph crosses at \((2,3)\).

Slant asymptote: \(y=x+1\). Crossing point: \((2,3)\).

- What is left over after dividing the highest-degree terms and subtracting their product? - Does the subtraction leave another term that belongs in the quotient? - Which entire polynomial quotient should define the slant line?

1. Dividing just the leading terms gives only the first quotient term, not the whole quotient. 2. Polynomial division gives \(x^2+1=(x+1)(x-1)+2\), so \(f(x)=x-1+\frac{2}{x+1}\). 3. The fraction becomes small for distant inputs, so the correct slant asymptote is \(y=x-1\). The student's line misses the constant term of the quotient.

The student stopped the division too early. The correct slant asymptote is \(y=x-1\).

- Find the zeros of the denominator and check them in the numerator. - Compare the degrees to decide what type of end-behavior asymptote is possible. - Divide to separate a linear part from a fraction with smaller numerator degree. - When would the fraction's value be zero?

1. The denominator is zero at \(x=-1\) and \(x=2\). The numerator is nonzero at both, so the vertical asymptotes are \(x=-1\) and \(x=2\). 2. Since \((x+1)(x-2)=x^2-x-2\), division gives \(x^3-1=(x+1)(x^2-x-2)+(3x+1)\). 3. Thus \(f(x)=x+1+\frac{3x+1}{x^2-x-2}\). The proper fraction approaches \(0\) for large positive or negative inputs, so the slant asymptote is \(y=x+1\). 4. The graph meets the slant asymptote when \(3x+1=0\). This gives \(x=-\frac{1}{3}\), which is allowed, and \(y=x+1=\frac{2}{3}\).

a) \(x=-1\) and \(x=2\) b) \(f(x)=x+1+\frac{3x+1}{x^2-x-2}\); slant asymptote: \(y=x+1\) c) Yes, at \(\left(-\frac{1}{3},\frac{2}{3}\right)\).

- First find where the function is undefined. - Compare the degrees of the numerator and denominator. - Use polynomial division to separate a linear part from a proper rational remainder. - Which term approaches zero as \(|x|\to\infty\)?

1. The denominator is zero at \(x=-2\). The numerator there is \(2(-2)^2+5(-2)-1=-3\neq 0\), so the vertical asymptote is \(x=-2\). 2. The numerator’s degree is one greater than the denominator’s degree, so use polynomial division. 3. Dividing gives \(2x^2+5x-1=(x+2)(2x+1)-3\). 4. Therefore, \(h(x)=2x+1-\frac{3}{x+2}\). 5. Since \(-\frac{3}{x+2}\to 0\) as \(|x|\to\infty\), the slant asymptote is \(y=2x+1\).

Vertical asymptote: \(x=-2\) Slant asymptote: \(y=2x+1\)

- Could polynomial division separate a linear expression from a remaining fraction? - How does the remaining fraction behave for large positive inputs? - Compare the function's value with the proposed line's value at the same input.

1. Divide to get \(f(x)=2x+3-\frac{3}{x+1}\). 2. As \(x\) becomes very large and positive, \(-\frac{3}{x+1}\) approaches \(0\). 3. The difference between the function and \(2x+3\) approaches \(0\), so Callum is correct. The slant asymptote is \(y=2x+3\).

Yes. \(f(x)=2x+3-\frac{3}{x+1}\), and the fractional term approaches \(0\) as \(x\) gets larger. Slant asymptote: \(y=2x+3\).

- Compare the degrees of each numerator and denominator. - What happens when both polynomials have the same degree? - When the numerator has degree one greater than the denominator, what does division tell you? - For the slant line, compare its direction on the left and the right.

1. For \(f\), numerator and denominator both have degree \(2\). The ratio of leading coefficients is \(\frac{10}{2}=5\), so the y-values approach \(5\) at both ends. Horizontal asymptote: \(y=5\). 2. Divide to obtain \(g(x)=x+1-\frac{2}{x+1}\). The fraction approaches \(0\) at both ends, so the slant asymptote is \(y=x+1\). The graph rises without bound to the right and falls without bound to the left. 3. For \(h\), the leading-coefficient ratio is \(\frac{-4}{2}=-2\). The y-values approach \(-2\) at both ends. Horizontal asymptote: \(y=-2\).

a) Approaches \(5\) at both ends; horizontal asymptote \(y=5\) b) Rises without bound as \(x\) gets very large and falls without bound as \(x\) gets very negative; slant asymptote \(y=x+1\) c) Approaches \(-2\) at both ends; horizontal asymptote \(y=-2\)

- Which input is excluded by the denominator? - Use \(x=0\) for the y-intercept and an output of zero for x-intercepts. - Can division rewrite the fraction as a line plus a smaller rational part? - Compare the graph with the line when the remaining fraction gets close to zero.

1. The denominator is zero at \(x=-1\), so the domain is \(\mathbb{R}\setminus\{-1\}\). 2. At \(x=0\), \(f(0)=4\), giving the y-intercept \((0,4)\). 3. The numerator equals zero at \(x=2\), giving the x-intercept \((2,0)\). 4. Expand and divide: \(\frac{x^2-4x+4}{x+1}=x-5+\frac{9}{x+1}\). 5. As \(x\) becomes very large or very negative, \(\frac{9}{x+1}\) approaches \(0\). The graph therefore approaches the line \(y=x-5\).

a) Domain: \(\mathbb{R}\setminus\{-1\}\); x-intercept: \((2,0)\); y-intercept: \((0,4)\) b) \(f(x)=x-5+\frac{9}{x+1}\); slant asymptote: \(y=x-5\). The fraction approaches \(0\) at both ends.

- What line do you get from polynomial division? - What happens to the remainder fraction for large positive and negative inputs? - Check both the direction of the line and the sign of the remainder on each end.

1. Divide \(2x^2-5x+1\) by \(x-2\). The quotient is \(2x-1\), with remainder \(-1\). 2. Thus \(k(x)=2x-1-\frac{1}{x-2}\). 3. The fractional part approaches \(0\) at both ends, so the graph approaches \(y=2x-1\). 4. As \(x\) becomes very large, \(k(x)\) rises without bound and lies below the slant line. As \(x\) becomes very negative, \(k(x)\) falls without bound and lies above the slant line.

\(k(x)=2x-1-\frac{1}{x-2}\); slant asymptote: \(y=2x-1\). Right end: rises without bound, approaching the line from below. Left end: falls without bound, approaching the line from above.

- Divide each term in the numerator by \(x\). - To find intersections of two graphs, set their equations equal. - Can a fraction with a nonzero constant numerator equal \(0\)?

1. Divide each term in the numerator by \(x\): \(g(x)=-x+6+\frac{4}{x}\). 2. Since \(\frac{4}{x}\to 0\) as \(x\to\pm\infty\), the slant asymptote is \(y=-x+6\). 3. To check for an intersection, solve \(-x+6+\frac{4}{x}=-x+6\). 4. This simplifies to \(\frac{4}{x}=0\), which has no solution. 5. Therefore, the graph does not intersect its slant asymptote.

a) \(g(x)=-x+6+\frac{4}{x}\); slant asymptote: \(y=-x+6\) b) The graph does not intersect the asymptote because \(\frac{4}{x}=0\) has no solution.

- Use polynomial division to write the function as a linear expression plus a remainder term. - The linear expression gives the slant asymptote. - Determine the sign of the remainder term for large positive \(x\).

1. Polynomial division gives \(f(x)=x-1+\frac{2}{x-3}\). 2. Therefore, the slant asymptote is \(y=x-1\). 3. For sufficiently large positive \(x\), \(x-3>0\), so \(\frac{2}{x-3}>0\). 4. Thus, \(f(x)>x-1\), and the graph approaches the asymptote from above as \(x\to\infty\).

1) The slant asymptote is \(y=x-1\). 2) The graph lies above the asymptote for large positive \(x\) because \(\frac{2}{x-3}>0\).

- What does division tell you that the leading-coefficient ratio does not? - A line needs both a slope and a constant term. - Check the remainder after division: does it become small at both ends?

1. Polynomial division gives \(g(x)=-2x+2+\frac{1}{x+1}\). 2. The leading-coefficient ratio correctly gives the slope \(-2\), but not the constant term of the line. 3. The fractional part approaches \(0\) at both ends, so the slant asymptote is \(y=-2x+2\), not \(y=-2x\).

Cem found the slope but omitted the constant term. Since \(g(x)=-2x+2+\frac{1}{x+1}\), the slant asymptote is \(y=-2x+2\).

- Write the function as the slant-asymptote expression plus a term that approaches zero. - Use a denominator factor to create the vertical asymptote. - Substitute the given point to determine the unknown constant.

1. Use the form \(f(x)=2x-4+\frac{a}{x-3}\). The rational term approaches \(0\) as \(|x|\to\infty\) and creates a vertical asymptote at \(x=3\). 2. Substitute \(P=(4, 6)\): \(6=2\cdot 4-4+\frac{a}{4-3}=4+a\). 3. Therefore, \(a=2\), so \(f(x)=2x-4+\frac{2}{x-3}\).

\(f(x)=2x-4+\frac{2}{x-3}\)

- Find the value that makes the denominator \(0\). - What happens to the rational remainder as \(\lvert x\rvert\) becomes large? - The y-intercept occurs when \(x=0\).

1. The denominator is zero at \(x=2\), so the domain is \(\mathbb{R}\setminus\{2\}\), and the vertical asymptote is \(x=2\). 2. Since \(\frac{6}{x-2}\to 0\) as \(x\to\pm\infty\), the slant asymptote is \(y=1.5x-3\). 3. Therefore, \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\). 4. Evaluate \(f(0)=-3+\frac{6}{-2}=-6\), so the y-intercept is \((0, -6)\).

a) Domain: \(\mathbb{R}\setminus\{2\}\); vertical asymptote: \(x=2\) b) \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\); slant asymptote: \(y=1.5x-3\) c) y-intercept: \((0, -6)\)

- Exclude values that make the denominator \(0\). - Use polynomial division to reveal the slant asymptote. - Set the numerator equal to \(0\) and verify that the solutions are in the domain.

1. The denominator is zero at \(x=3\), so the domain is \(\mathbb{R}\setminus\{3\}\). 2. Polynomial division gives \(g(x)=x-2-\frac{2}{x-3}\). Therefore, the slant asymptote is \(y=x-2\). 3. Factor the numerator: \(x^2-5x+4=(x-1)(x-4)\). Neither zero is excluded from the domain, so the x-intercepts are \((1, 0)\) and \((4, 0)\).

a) \(\mathbb{R}\setminus\{3\}\) b) \(y=x-2\) c) \((1, 0)\) and \((4, 0)\)

- Find the value that makes the denominator \(0\). - Compare \(f_k(-x)\) with \(f_k(x)\). - Divide each numerator term by \(3x\) to reveal the slant asymptote. - Substitute the coordinates of the given point.

1. The denominator is zero at \(x=0\), so the domain is \(\mathbb{R}\setminus\{0\}\). 2. Since \(f_k(-x)=-f_k(x)\), each graph is symmetric about the origin. 3. At \(x=0\), the numerator is \(-9\), so the vertical asymptote is \(x=0\). 4. Rewrite \(f_k(x)=\frac{k}{3}x-\frac{3}{x}\). Since \(-\frac{3}{x}\to 0\), the slant asymptote is \(y=\frac{k}{3}x\). 5. Use the point \((3, 2)\): \(2=\frac{9k-9}{9}=k-1\), so \(k=3\).

a) Domain: \(\mathbb{R}\setminus\{0\}\); symmetric about the origin b) Vertical asymptote: \(x=0\); slant asymptote: \(y=\frac{k}{3}x\) c) \(k=3\)

- What happens to the numerator and denominator at the excluded input? - Rewrite the function as a linear expression plus a fraction. - Match both the slope and constant term with the proposed slant line. - When \(a=0\), which part gets small far from the origin?

1. At \(x=0\), the denominator is zero but the numerator is \(4\). Thus \(x=0\) is always a vertical asymptote. 2. Split the fraction: \(g_{a,b}(x)=\frac{a}{2}x+\frac{b}{2}+\frac{2}{x}\). 3. The final term approaches \(0\) for large positive or negative inputs. Matching the line \(y=1.5x+2\) gives \(\frac{a}{2}=1.5\) and \(\frac{b}{2}=2\), so \(a=3\) and \(b=4\). 4. With \(a=0\), the function is \(g_{0,b}(x)=\frac{b}{2}+\frac{2}{x}\). The fractional part approaches \(0\), so the horizontal asymptote is \(y=\frac{b}{2}\).

a) At \(x=0\), the denominator is zero and the numerator is nonzero. b) \(a=3\), \(b=4\) c) \(y=\frac{b}{2}\), because \(\frac{2}{x}\) approaches \(0\) for very large positive or negative inputs.

- Find where the denominator is \(0\) and where the numerator is \(0\). - Substitute \(x=0\) for the y-intercept. - Use polynomial division to identify the slant asymptote. - Separate the direction of the end behavior from the vertical position of the asymptote.

1. The denominator is zero at \(x=-k\), so the domain is \(\mathbb{R}\setminus\{-k\}\). 2. Solve \(x^2-4k^2=0\), giving \(x=2k\) and \(x=-2k\). Because \(k\neq 0\), neither value equals \(-k\). The x-intercepts are \((2k, 0)\) and \((-2k, 0)\). 3. Evaluate \(f_k(0)=\frac{-4k^2}{k}=-4k\), so the y-intercept is \((0, -4k)\). 4. Polynomial division gives \(f_k(x)=x-k-\frac{3k^2}{x+k}\). Thus, the slant asymptote is \(y=x-k\). 5. For every fixed nonzero \(k\), the remainder term approaches \(0\). Therefore, \(f_k(x)\to\infty\) as \(x\to\infty\), and \(f_k(x)\to-\infty\) as \(x\to-\infty\). These directions are independent of \(k\), although the asymptote's y-intercept depends on \(k\).

a) Domain: \(\mathbb{R}\setminus\{-k\}\); x-intercepts: \((2k, 0)\) and \((-2k, 0)\) b) y-intercept: \((0, -4k)\) c) \(f_k(x)\to\infty\) as \(x\to\infty\), and \(f_k(x)\to-\infty\) as \(x\to-\infty\); slant asymptote: \(y=x-k\)

- Use polynomial division or rewrite the numerator in terms of \(x-4\). - Set the function equal to its asymptote to test for intersections. - Examine the sign of the remainder term on each side of \(x=4\).

1. Rewrite the function as \(h(x)=x+\frac{7}{x-4}\). 2. Since \(\frac{7}{x-4}\to 0\) as \(x\to\pm\infty\), the slant asymptote is \(y=x\). 3. To find an intersection, solve \(x+\frac{7}{x-4}=x\). This reduces to \(\frac{7}{x-4}=0\), which has no solution. 4. For \(x>4\), \(\frac{7}{x-4}>0\), so the graph lies above \(y=x\). For \(x<4\), \(\frac{7}{x-4}<0\), so the graph lies below \(y=x\).

The slant asymptote is \(y=x\). The equation \(\frac{7}{x-4}=0\) has no solution, so the graph never intersects the asymptote. It lies above the asymptote for \(x>4\) and below it for \(x<4\).

- Find the denominator zero and evaluate the numerator there. - Use polynomial division. - The graph meets its asymptote only when the remainder term is \(0\). - Set the numerator equal to \(0\) for x-intercepts.

1. The denominator is zero at \(x=2\), while the numerator equals \(1\), so the domain is \(\mathbb{R}\setminus\{2\}\), and the vertical asymptote is \(x=2\). 2. Polynomial division gives \(h(x)=2x+1+\frac{1}{x-2}\). Thus, the slant asymptote is \(y=2x+1\). 3. An intersection with the asymptote would require \(\frac{1}{x-2}=0\), which has no solution. 4. X-intercepts satisfy \(2x^2-3x-1=0\). The quadratic formula gives \(x=\frac{3\pm\sqrt{17}}{4}\).

a) Domain: \(\mathbb{R}\setminus\{2\}\); vertical asymptote: \(x=2\) b) \(h(x)=2x+1+\frac{1}{x-2}\); slant asymptote: \(y=2x+1\) c) No intersection with the slant asymptote d) x-intercepts: \(\left(\frac{3-\sqrt{17}}{4}, 0\right)\) and \(\left(\frac{3+\sqrt{17}}{4}, 0\right)\)

- Rewrite each function as a linear expression plus a proper rational remainder. - Use term-by-term division when the denominator is a monomial. - Use polynomial division when necessary. - Which part remains after the term that approaches \(0\) is removed?

1. The function \(g_1\) is already written as \(x-2\) plus a term that approaches \(0\), so its slant asymptote is \(y=x-2\). 2. Divide term by term: \(g_2(x)=x-2+\frac{1}{x}\). The remainder term approaches \(0\), so the asymptote is \(y=x-2\). 3. Similarly, \(g_3(x)=x-2+\frac{3}{2x}\). The remainder term approaches \(0\), so the asymptote is \(y=x-2\). 4. Polynomial division gives \(g_4(x)=x-2+\frac{5}{x+2}\). The remainder term approaches \(0\), so the asymptote is \(y=x-2\). 5. Therefore, all four graphs have the same slant asymptote.

All four functions have the slant asymptote \(y=x-2\).

- Which value makes the denominator equal to \(0\)? - What happens to the rational remainder as \(|x|\) becomes large? - To test for an intersection, set the function equal to the asymptote.

1. The denominator requires \(x+3\neq 0\), so the domain is \(\mathbb{R}\setminus\{-3\}\). The vertical asymptote is \(x=-3\). 2. Since \(-\frac{4}{x+3}\to 0\) as \(x\to\pm\infty\), the graph approaches \(y=\frac{1}{2}x-1\). 3. Thus, \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\). The slant asymptote is \(y=\frac{1}{2}x-1\). 4. Setting the function equal to the asymptote gives \(-\frac{4}{x+3}=0\), which has no solution. Therefore, the graph does not intersect the slant asymptote.

a) Domain: \(\mathbb{R}\setminus\{-3\}\); vertical asymptote: \(x=-3\) b) \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\); slant asymptote: \(y=\frac{1}{2}x-1\) c) The graph does not intersect the slant asymptote.

- Use the asymptotes to identify \(b\) and \(c\). - Substitute the given point to find \(a\). - Set the full function equal to zero. - Rewrite \(m\) as a linear expression plus a rational term that approaches zero.

1. The vertical asymptote gives \(b=2\), and the horizontal asymptote gives \(c=3\). 2. Substitute \(P=(3, 5)\): \(5=\frac{a}{3-2}+3\), so \(a=2\). Thus, \(k(x)=\frac{2}{x-2}+3\). 3. To find the zero, solve \(\frac{2}{x-2}+3=0\). Then \(2=-3(x-2)\), so \(3x=4\) and \(x=\frac{4}{3}\). 4. \(m(x)=\frac{2}{x-2}+3-3x=-3x+3+\frac{2}{x-2}\). 5. The vertical asymptote remains \(x=2\). Since \(\frac{2}{x-2}\to0\) as \(|x|\to\infty\), the slant asymptote is \(y=-3x+3\).

a) \(a=2\), \(b=2\), \(c=3\) b) \(x=\frac{4}{3}\) c) Vertical asymptote: \(x=2\); slant asymptote: \(y=-3x+3\)

- The numerator's degree is one greater than the denominator's degree. - Use polynomial division to separate a linear quotient from a proper rational remainder. - What happens to the remainder term as \(|x|\) becomes large?

1. Divide \(0.5x^2+2x-1\) by \(x-2\): \(k(x)=0.5x+3+\frac{5}{x-2}\). 2. Since \(\frac{5}{x-2}\to 0\) as \(x\to\pm\infty\), the graph approaches \(y=0.5x+3\). 3. Therefore, \(k(x)\to\infty\) as \(x\to\infty\), and \(k(x)\to-\infty\) as \(x\to-\infty\).

1) \(k(x)\to\infty\) as \(x\to\infty\), and \(k(x)\to-\infty\) as \(x\to-\infty\) 2) Slant asymptote: \(y=0.5x+3\)

- Find the denominator zero and the numerator zeros. - Use polynomial division. - The multiplicity of the denominator factor helps determine the sign behavior at the vertical asymptote. - Test \(f(-x)=f(x)\) and \(f(-x)=-f(x)\).

1. The denominator is zero at \(x=1\), so the domain is \(\mathbb{R}\setminus\{1\}\). 2. Solve \(x^2-4=0\), giving x-intercepts \((-2, 0)\) and \((2, 0)\). Also, \(f(0)=4\), so the y-intercept is \((0, 4)\). 3. Polynomial division gives \(f(x)=x+1-\frac{3}{x-1}\). 4. The vertical asymptote is \(x=1\). Because the denominator factor has odd multiplicity and the numerator is nonzero at \(x=1\), the function changes sign across the asymptote. The slant asymptote is \(y=x+1\). 5. Since \(f(-x)\neq f(x)\) and \(f(-x)\neq-f(x)\), the graph is symmetric about neither the y-axis nor the origin.

a) Domain: \(\mathbb{R}\setminus\{1\}\); x-intercepts: \((-2, 0)\), \((2, 0)\); y-intercept: \((0, 4)\) b) \(f(x)=x+1-\frac{3}{x-1}\) c) Vertical asymptote: \(x=1\), with a sign change; slant asymptote: \(y=x+1\) d) No symmetry about the y-axis or the origin

- Could the numerator ever equal zero for a real input? - Divide the two numerator terms separately by \(2x\). - Compare the function values at opposite inputs. - What sign does the denominator have just to the left and right of zero?

1. The denominator is zero at \(x=0\), so the domain is \(\mathbb{R}\setminus\{0\}\). Since \(x^2+3>0\) for every real input, there are no x-intercepts. 2. The numerator is nonzero at \(x=0\), giving vertical asymptote \(x=0\). Rewrite \(g(x)=\frac{1}{2}x+\frac{3}{2x}\). The final term approaches \(0\) at both ends, so the slant asymptote is \(y=\frac{1}{2}x\). 3. Because \(g(-x)=-g(x)\), the graph is symmetric about the origin. 4. Near \(0\), the numerator stays positive. The denominator is positive just to the right and negative just to the left, so the y-values rise without bound from the right and fall without bound from the left.

a) \(\mathbb{R}\setminus\{0\}\); no x-intercepts b) Vertical asymptote: \(x=0\); slant asymptote: \(y=\frac{1}{2}x\) c) Symmetric about the origin d) Approaching \(0\) from the right: rises without bound; approaching from the left: falls without bound.

- A rational function is undefined where its denominator is \(0\). - Check whether the excluded factor cancels before identifying a vertical asymptote. - Use polynomial division to find a slant asymptote. - Substitute \(x=4\) to evaluate the function.

1. The denominator is zero at \(x=2\), so the domain is \(\mathbb{R}\setminus\{2\}\). 2. The numerator is \(-2\) at \(x=2\), so the factor does not cancel. Therefore, the vertical asymptote is \(x=2\). 3. Polynomial division gives \(f(x)=x-1-\frac{2}{x-2}\). Since the remainder term approaches \(0\), the slant asymptote is \(y=x-1\). 4. Evaluate \(f(4)=\frac{4^2-3\cdot 4}{4-2}=\frac{4}{2}=2\).

a) \(\mathbb{R}\setminus\{2\}\) b) Vertical asymptote: \(x=2\); slant asymptote: \(y=x-1\) c) \(f(4)=2\)

- Divide each term in the numerator by \(x\). - Which term determines the overall end behavior, and which term approaches \(0\)? - Compare \(g(x)\) directly with the linear part by checking the sign of their difference.

1. Divide each term in the numerator by \(x\): \(g(x)=2x+3-\frac{4}{x}\). 2. Since \(-\frac{4}{x}\to 0\), the slant asymptote is \(y=2x+3\). Therefore, \(g(x)\to\infty\) as \(x\to\infty\), and \(g(x)\to-\infty\) as \(x\to-\infty\). 3. For positive \(x\), \(-\frac{4}{x}<0\). Thus, \(g(x)<2x+3\), so the graph approaches the asymptote from below as \(x\to\infty\).

1) \(g(x)=2x+3-\frac{4}{x}\) 2) \(g(x)\to\infty\) as \(x\to\infty\), and \(g(x)\to-\infty\) as \(x\to-\infty\) 3) The graph lies below the slant asymptote \(y=2x+3\) for large positive \(x\).

- Use polynomial division to separate a linear quotient and a remainder term. - Which part approaches \(0\) as \(x\to\infty\)? - The sign of the remainder term determines whether the graph is above or below the asymptote.

1. Polynomial division gives \(f(x)=x-1-\frac{2}{x+3}\). 2. Since \(-\frac{2}{x+3}\to 0\) as \(x\to\pm\infty\), the slant asymptote is \(y=x-1\). 3. For large positive \(x\), \(-\frac{2}{x+3}<0\). Therefore, \(f(x)<x-1\), so the graph approaches the asymptote from below.

1) \(f(x)=x-1-\frac{2}{x+3}\) 2) \(y=x-1\) 3) The graph approaches the asymptote from below as \(x\to\infty\).

- In polynomial division, the remainder must have lower degree than the divisor. - The graph meets the asymptote when the remainder term equals \(0\). - When is a rational expression equal to \(0\)?

1. Polynomial division gives \(g(x)=2x-1+\frac{-2x+3}{x^2+2}\). 2. Since the remainder term approaches \(0\) as \(x\to\pm\infty\), the slant asymptote is \(y=2x-1\). 3. The graph intersects the asymptote when \(\frac{-2x+3}{x^2+2}=0\). Because \(x^2+2>0\), solve \(-2x+3=0\), giving \(x=\frac{3}{2}\). 4. Substitute into the asymptote: \(y=2\cdot\frac{3}{2}-1=2\). The intersection is \(\left(\frac{3}{2}, 2\right)\).

Slant asymptote: \(y=2x-1\) Intersection: \(\left(\frac{3}{2}, 2\right)\)

- Use polynomial division to identify the linear part and the remainder term. - The vertical distance from the function to the asymptote is the absolute value of the remainder term. - Solve the resulting absolute-value inequality carefully.

1. Polynomial division gives \(f(x)=3x-5+\frac{8}{x+2}\). Therefore, the slant asymptote is \(a(x)=3x-5\). 2. The required condition is \(\lvert f(x)-a(x)\rvert<0.4\), so \(\left\lvert\frac{8}{x+2}\right\rvert<0.4\). 3. This is equivalent to \(\frac{8}{\lvert x+2\rvert}<0.4\), which gives \(\lvert x+2\rvert>20\). 4. Therefore, \(x+2>20\) or \(x+2<-20\), so \(x>18\) or \(x<-22\).

1) \(a(x)=3x-5\) 2) \(x\in(-\infty, -22)\cup(18, \infty)\)

- Use polynomial division to separate a linear part and a proper rational remainder. - The sign of \(f(x)-g(x)\) determines the graph's position relative to \(g\). - Check the sign of the remainder term for large positive and large negative values of \(x\).

1. Polynomial division gives \(f(x)=3x+\frac{2}{x-2}\). 2. The linear approximation, which is also the slant asymptote, is \(g(x)=3x\). 3. The difference is \(f(x)-g(x)=\frac{2}{x-2}\). 4. As \(x\to\infty\), the difference is positive, so the graph approaches \(g\) from above. 5. As \(x\to-\infty\), the difference is negative, so the graph approaches \(g\) from below.

\(g(x)=3x\) As \(x\to\infty\), the graph approaches from above. As \(x\to-\infty\), the graph approaches from below.

- Use polynomial division when the numerator degree is one greater. - Combine the origin-symmetry condition with the two end limits. - Recall how even and odd functions behave when evaluated at \(-x\).

1. a) True. Polynomial division produces a nonconstant linear quotient plus a remainder term whose ratio to \(q(x)\) approaches \(0\). The linear quotient is the slant asymptote. 2. b) False. Origin symmetry means \(f(-x)=-f(x)\). If both end limits are the same finite number \(L\), this symmetry requires \(L=-L\), so \(L=0\). Therefore, \(y=2\) cannot be the horizontal asymptote. 3. c) True. A polynomial with only even powers is even, so \(p(-x)=p(x)\). A polynomial with only odd powers is odd, so \(q(-x)=-q(x)\). Hence \(f(-x)=\frac{p(x)}{-q(x)}=-f(x)\) wherever the function is defined.

a) True b) False c) True

- Start with the equation of the slant line. - What kind of rational term can become small at both ends without ever becoming undefined? - Use the y-intercept to determine any remaining coefficient.

1. Choose a denominator with no real zeros, such as \(x^2+1\). 2. Set \(g(x)=-x+1+\frac{a}{x^2+1}\). The fractional part approaches \(0\) far to the left and right, so the slant asymptote is \(y=-x+1\). 3. Use \(g(0)=2\): \(1+a=2\), so \(a=1\). 4. Thus one possible function is \(g(x)=-x+1+\frac{1}{x^2+1}\).

One possible function: \(g(x)=-x+1+\frac{1}{x^2+1}\).

- Where does this function stop having a defined value? - What happens to the fraction when \(x\) is very close to that excluded input? - How could you compare the full rule with a simpler expression for very large inputs?

1. The denominator of \(\frac{2}{x}\) is zero at \(x=0\). The fraction grows without bound in magnitude as \(x\) gets close to \(0\), so the vertical asymptote is \(x=0\). 2. Compare \(f(x)\) with the line \(y=0.5x\): the difference is \(f(x)-0.5x=\frac{2}{x}\). That difference approaches \(0\) as \(x\) becomes very large positive or negative, so the slant asymptote is \(y=0.5x\).

Vertical asymptote: \(x=0\), because the fraction \(\frac{2}{x}\) is undefined and unbounded there. Slant asymptote: \(y=0.5x\), because the difference \(\frac{2}{x}\) approaches \(0\) for large positive or negative \(x\).

- Write the function as the slant-asymptote expression plus a rational remainder. - Use the denominator to create the vertical asymptote. - Substitute the given point to find the unknown constant.

1. Use the form \(f(x)=2x-3+\frac{k}{x-2}\). 2. Substitute \(P=(1,0)\): \(0=2\cdot1-3+\frac{k}{1-2}=-1-k\). Therefore, \(k=-1\). 3. One possible function is \(f(x)=2x-3-\frac{1}{x-2}\). 4. As one fraction, \(f(x)=\frac{(2x-3)(x-2)-1}{x-2}=\frac{2x^2-7x+5}{x-2}\).

One possible formula is \(f(x)=2x-3-\frac{1}{x-2}\).

- Read two convenient positions on the dashed line to find its slope. - Which input is excluded at the vertical asymptote? - Choose a clear point on the curved graph and substitute it into a trial function.

1. The dashed line passes through the origin with slope \(1\), giving slant asymptote \(y=x\). 2. The curve has vertical asymptote \(x=0\), so try \(f(x)=x+\frac{k}{x}\). 3. A readable point on the graph is \((1,3)\). Substitute: \(3=1+k\), giving \(k=2\). 4. Therefore one possible rule is \(f(x)=x+\frac{2}{x}\).

Slant asymptote: \(y=x\); one possible rule: \(f(x)=x+\frac{2}{x}\).

- Check when the original denominator vanishes. - Separate a numerator made from a product and an added term. - Consider which part becomes small far from the excluded input, and whether it can ever equal zero.

1. The denominator is zero at \(x=1\), so \(x=1\) is excluded. 2. Separate the numerator: \(f(x)=2x+3+\frac{4}{x-1}\) for \(x\neq1\). 3. The fraction becomes small for inputs far from \(1\), so the slant asymptote is \(y=2x+3\). 4. Because \(\frac{4}{x-1}\) never equals zero on its domain, the graph cannot cross this line.

a) \(x=1\). b) \(f(x)=2x+3+\frac{4}{x-1}\), with \(x\neq1\). c) \(y=2x+3\); it is never crossed.

- A rectangle's perimeter is twice the sum of its length and width. - Identify a straight-line part and an additional term that becomes small for large widths. - When solving for the target perimeter, use the positive domain before multiplying by \(x\).

1. The perimeter is twice the sum of the side lengths: \(P(x)=2x+\frac{144}{x}\). 2. The difference from \(2x\) is \(\frac{144}{x}\), which becomes small as \(x\) grows. Thus, the slant asymptote is \(y=2x\). 3. Solve \(2x+\frac{144}{x}=36\). Since \(x>0\), multiply by \(x\): \(2x^2-36x+144=0\), or \((x-6)(x-12)=0\). 4. The dimensions are \(6\,\text{ft}\) by \(12\,\text{ft}\), in either order.

a) \(P(x)=2x+\frac{144}{x}\). b) \(y=2x\); the extra perimeter contribution \(\frac{144}{x}\) shrinks as \(x\) grows. c) \(6\,\text{ft}\) by \(12\,\text{ft}\).

- Analyze each function separately. - Check excluded values when looking for vertical asymptotes. - Use polynomial division when the numerator's degree is one greater than the denominator's degree. - Compare the leading terms to determine end behavior.

1. For \(h\), the denominator has greater degree than the numerator, so \(h(x)\to 0\) as \(x\to\infty\). Thus, a) is true. 2. Polynomial division gives \(g(x)=x+1+\frac{2}{x-1}\). Therefore, \(y=x+1\) is the slant asymptote, so b) is true. 3. The graph of \(f\) has a vertical asymptote at \(x=-3\) and a horizontal asymptote at \(y=-2\), for a total of two. Since \(x^2+2>0\) for all real \(x\), \(h\) has no vertical asymptote and has only the horizontal asymptote \(y=0\). Therefore, c) is false. 4. As \(x\to\infty\), \(f(x)\to-2\), \(g(x)\to\infty\), and \(h(x)\to 0\). Therefore, d) is false.

a) True b) True c) False; \(f\) has two asymptotes, while \(h\) has one. d) False; as \(x\to\infty\), \(f(x)\to-2\), \(g(x)\to\infty\), and \(h(x)\to 0\).

- Symmetry about the origin requires \(g(-x)=-g(x)\). - Use denominator factors for the two vertical asymptotes. - Use polynomial division to control the slant asymptote. - Choose an odd numerator and an even denominator.

1. The vertical asymptotes suggest the denominator \((x-2)(x+2)=x^2-4\). 2. Symmetry about the origin requires an odd function. Since the denominator is even, choose an odd numerator. 3. A slant asymptote \(y=x\) requires the numerator degree to be one greater than the denominator degree and the quotient from polynomial division to be \(x\). 4. Choose \(g(x)=\frac{x^3}{x^2-4}\). Polynomial division gives \(g(x)=x+\frac{4x}{x^2-4}\), so the slant asymptote is \(y=x\). Also, \(g(-x)=-g(x)\), and the numerator is nonzero at \(x=\pm2\).

One possible formula is \(g(x)=\frac{x^3}{x^2-4}\).

- Write the algebraic expression produced by a horizontal shift of the first function. - Compare that shifted expression term by term with \(g(x)\). - Determine whether an additional vertical change is needed after the horizontal shift.

1. The denominator in \(f\) is zero at \(x=0\), so the domain is \(\mathbb{R}\setminus\{0\}\), and the vertical asymptote is \(x=0\). 2. Since \(\frac{4}{x}\to0\) as \(x\to\pm\infty\), the slant asymptote is \(y=x\). 3. A shift \(2\) units to the right gives \(f(x-2)=(x-2)+\frac{4}{x-2}=x-2+\frac{4}{x-2}\). 4. Compare this with \(g(x)=x+\frac{4}{x-2}\). Since \(g(x)=f(x-2)+2\), the graph must also be shifted \(2\) units up. 5. Therefore, the claim is false, and the translation vector is \((2,2)\).

a) Domain: \(\mathbb{R}\setminus\{0\}\); vertical asymptote: \(x=0\); slant asymptote: \(y=x\) b) The claim is false. The translation vector is \((2,2)\).

- Use an even denominator multiplicity to avoid a sign change. - Write the function as the slant-asymptote expression plus a rational term that approaches zero. - Make sure the added rational term is nonzero near \(x=4\). - You may leave the answer in sum form or combine it into one fraction.

1. A vertical asymptote at \(x=4\) without a sign change can be created with the squared denominator factor \((x-4)^2\). 2. Write the function as its slant asymptote plus a proper rational term: \(f(x)=2x-1+\frac{k}{(x-4)^2}\), where \(k\ne0\). 3. Choose \(k=1\). Then \(f(x)=2x-1+\frac{1}{(x-4)^2}\). 4. As one fraction, \(f(x)=\frac{(2x-1)(x-4)^2+1}{(x-4)^2}=\frac{2x^3-17x^2+40x-15}{x^2-8x+16}\). 5. The rational remainder approaches \(0\) as \(|x|\to\infty\), and the numerator is \(1\) at \(x=4\), so the required asymptotes remain.

One possible formula is \(f(x)=2x-1+\frac{1}{(x-4)^2}\).

- Find values that make the denominator equal to \(0\). - Use polynomial division to separate a linear part and a remainder. - In part b), rewrite the rational term before combining it with the linear term.

1. For a), the denominator is zero at \(x=-1\), and the numerator is nonzero there, so the vertical asymptote is \(x=-1\). Polynomial division gives \(g(x)=x-1+\frac{4}{x+1}\), so the slant asymptote is \(y=x-1\). 2. For b), the denominator is zero at \(x=2\), and the numerator of the fractional term is nonzero there, so the vertical asymptote is \(x=2\). Rewrite \(\frac{2x}{x-2}=2+\frac{4}{x-2}\). Then \(g(x)=\frac{1}{3}x-2-\frac{4}{x-2}\), so the slant asymptote is \(y=\frac{1}{3}x-2\).

a) Vertical asymptote: \(x=-1\); slant asymptote: \(y=x-1\) b) Vertical asymptote: \(x=2\); slant asymptote: \(y=\frac{1}{3}x-2\)