Free math worksheets

- Compare the blue curve with each dashed line far to the left and far to the right. - A slant asymptote is the line whose distance from the curve shrinks as \(|x|\) becomes large.

1. Far to the left and far to the right, the blue curve gets closer to \(l_1\). 2. The curve does not approach \(l_2\); the gap between them stays visibly large. 3. Therefore, \(l_1\) is the slant asymptote.

\(l_1\). The curve approaches \(l_1\) as \(x\to\pm\infty\).

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

- 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.

- 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.

- First identify denominator zeros and check whether either factor cancels. - Compare the numerator and denominator degrees to decide what kind of end-behavior asymptote is possible. - Rewrite the rational function as a linear part plus a remainder term whose degree is lower than the denominator's degree. - Once you have that form, ask when the remainder term can be zero.

1. The denominator is zero at \(x=-1\) and \(x=2\). The numerator is nonzero at both values, so the vertical asymptotes are \(x=-1\) and \(x=2\). 2. Divide \(x^3-1\) by \(x^2-x-2\): \(f(x)=x+1+\frac{3x+1}{x^2-x-2}\). 3. The proper rational term approaches \(0\) for large positive or negative \(x\), so the slant asymptote is \(y=x+1\). 4. The graph meets the slant asymptote when the remainder term is \(0\). Thus \(3x+1=0\), so \(x=-\frac{1}{3}\). 5. On the asymptote, \(y=x+1=\frac{2}{3}\). Therefore, the intersection point is \(\left(-\frac{1}{3},\frac{2}{3}\right)\).

a) \(x=-1\) and \(x=2\) b) \(y=x+1\) c) Yes. The graph intersects the slant asymptote 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: \(y=2x+1\)

- Use polynomial division to separate a linear expression and a proper rational remainder. - What happens to the remainder term as \(x\to\infty\)? - The quotient from polynomial division determines the slant asymptote.

1. Use polynomial division: \(\frac{2x^2+5x}{x+1}=2x+3-\frac{3}{x+1}\). 2. As \(x\to\infty\), the remainder term \(-\frac{3}{x+1}\to0\). 3. Therefore, the difference between \(f(x)\) and \(2x+3\) approaches \(0\). 4. The prediction is correct, and the slant asymptote is \(y=2x+3\).

The prediction is correct. Since \(f(x)=2x+3-\frac{3}{x+1}\) and the remainder term approaches \(0\), the slant asymptote is \(y=2x+3\).

- Compare the degrees of the numerator and denominator. - When the degrees are equal, use the ratio of leading coefficients. - When the numerator’s degree is one greater, use polynomial division. - The remainder term determines how the graph approaches the polynomial quotient.

1. For part a, the numerator and denominator have the same degree. The ratio of leading coefficients is \(\frac{10}{2}=5\), so \(f(x)\to 5\) as \(x\to\pm\infty\). The horizontal asymptote is \(y=5\). 2. For part b, polynomial division gives \(g(x)=x+1-\frac{2}{x+1}\). The function is unbounded as \(x\to\pm\infty\), but the difference between \(g(x)\) and \(x+1\) approaches \(0\). The slant asymptote is \(y=x+1\). 3. For part c, the numerator and denominator have the same degree. The ratio of leading coefficients is \(\frac{-4}{2}=-2\), so \(h(x)\to-2\) as \(x\to\pm\infty\). The horizontal asymptote is \(y=-2\).

a) Approaches \(5\); horizontal asymptote \(y=5\) b) Is unbounded; slant asymptote \(y=x+1\) c) Approaches \(-2\); horizontal asymptote \(y=-2\)

- Exclude values that make the denominator \(0\). - Set \(x=0\) for the y-intercept and set the numerator equal to \(0\) for x-intercepts. - Use polynomial division to reveal the slant asymptote.

1. The denominator is zero at \(x=-1\), so the domain is \(\mathbb{R}\setminus\{-1\}\). 2. Evaluate \(f(0)=4\), so the y-intercept is \((0, 4)\). 3. The numerator is zero when \((x-2)^2=0\), so the x-intercept is \((2, 0)\). 4. Expanding and dividing gives \(\frac{x^2-4x+4}{x+1}=x-5+\frac{9}{x+1}\). 5. Since \(\frac{9}{x+1}\to 0\) as \(x\to\pm\infty\), the slant asymptote is \(y=x-5\).

a) Domain: \(\mathbb{R}\setminus\{-1\}\); y-intercept: \((0, 4)\); x-intercept: \((2, 0)\) b) Slant asymptote: \(y=x-5\)

- When the numerator's degree is exactly one greater than the denominator's degree, polynomial division can reveal a slant asymptote. - Separate the expression into a linear quotient and a proper rational remainder. - What happens to the remainder term as \(|x|\) becomes large?

1. Divide \(2x^2-5x+1\) by \(x-2\). 2. The first quotient term is \(2x\). Subtracting \(2x(x-2)=2x^2-4x\) leaves \(-x+1\). 3. The next quotient term is \(-1\). Subtracting \(-(x-2)=-x+2\) leaves a remainder of \(-1\). 4. Therefore, \(k(x)=2x-1-\frac{1}{x-2}\). 5. Since \(-\frac{1}{x-2}\to 0\) as \(x\to\pm\infty\), the graph approaches \(y=2x-1\).

\(k(x)=2x-1-\frac{1}{x-2}\). As \(x\to\pm\infty\), the graph approaches the slant asymptote \(y=2x-1\).

- 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 \(\frac{2}{x-3}>0\).

- Complete the polynomial division rather than comparing only the leading terms. - A line \(y=mx+b\) requires both a slope and a y-intercept. - Identify which part of the quotient the leading-coefficient comparison determines.

1. Divide \(-2x^2+3\) by \(x+1\): \(g(x)=-2x+2+\frac{1}{x+1}\). 2. The ratio of leading coefficients correctly gives the slope \(-2\), but it does not determine the constant term of the asymptote. 3. Since \(\frac{1}{x+1}\to0\) as \(x\to\pm\infty\), the correct slant asymptote is \(y=-2x+2\).

The claim is incorrect. Polynomial division gives \(g(x)=-2x+2+\frac{1}{x+1}\). Comparing leading coefficients gives only the slope; Cem omitted the constant term \(2\). 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) \(k=3\)

- Evaluate the numerator at the excluded x-value. - Split the rational expression into separate terms. - Match the slope and constant term with the given line. - When \(a=0\), identify the term that approaches \(0\).

1. The denominator is zero at \(x=0\), while the numerator equals \(4\). Therefore, \(x=0\) is always a vertical asymptote. 2. Rewrite \(g_{a,b}(x)=\frac{a}{2}x+\frac{b}{2}+\frac{2}{x}\). Thus, the slant asymptote is \(y=\frac{a}{2}x+\frac{b}{2}\). 3. Match coefficients with \(y=1.5x+2\): \(\frac{a}{2}=1.5\) and \(\frac{b}{2}=2\). Hence, \(a=3\) and \(b=4\). 4. If \(a=0\), then \(g_{0,b}(x)=\frac{b}{2}+\frac{2}{x}\). Therefore, \(\lim_{x\to\pm\infty}g_{0,b}(x)=\frac{b}{2}\), and the horizontal asymptote is \(y=\frac{b}{2}\).

a) \(x=0\) b) \(a=3\), \(b=4\) c) Horizontal asymptote: \(y=\frac{b}{2}\)

- 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

- Can \(x^2+3\) equal \(0\) for a real value of \(x\)? - Split the fraction into two terms. - Compare \(g(-x)\) with \(g(x)\). - Check the sign of the denominator on each side of \(0\).

1. The denominator is zero at \(x=0\), so the domain is \(\mathbb{R}\setminus\{0\}\). Since \(x^2+3>0\) for all real \(x\), there are no x-intercepts. 2. The vertical asymptote is \(x=0\). Rewrite \(g(x)=\frac{1}{2}x+\frac{3}{2x}\). Since the remainder term approaches \(0\), the slant asymptote is \(y=\frac{1}{2}x\). 3. Since \(g(-x)=-g(x)\), the graph is symmetric about the origin. 4. As \(x\to 0^+\), the numerator approaches \(3\) and the denominator approaches \(0\) through positive values, so \(g(x)\to\infty\). As \(x\to 0^-\), \(g(x)\to-\infty\).

a) Domain: \(\mathbb{R}\setminus\{0\}\); no x-intercepts b) Vertical asymptote: \(x=0\); slant asymptote: \(y=\frac{1}{2}x\) c) Symmetric about the origin d) \(\lim_{x\to 0^+}g(x)=\infty\), \(\lim_{x\to 0^-}g(x)=-\infty\)

- 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) \(f(4)=2\)

- Use the chain rule for the first derivative and then differentiate again. - An inflection point requires a change in concavity. - Factor \(x^2\) from under the square root when evaluating the limits. - To find each asymptote, evaluate the limit of the difference between \(f(x)\) and a line with the corresponding slope.

1. Differentiate: \(f'(x)=\frac{x}{\sqrt{x^2+5}}\) and \(f''(x)=\frac{5}{(x^2+5)^{3/2}}\). 2. Solve \(f'(x)=0\). This gives \(x=0\). Since \(f''(0)=\frac{1}{\sqrt{5}}>0\), the point is a local minimum. Its coordinates are \((0,\sqrt{5})\). 3. Because \(f''(x)>0\) for every real \(x\), the concavity never changes. Therefore, the graph has no inflection points. 4. Using \(\sqrt{x^2}=|x|\), \(\lim_{x\to\infty}\frac{x}{\sqrt{x^2+5}}=1\) and \(\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+5}}=-1\). 5. For the right-hand asymptote, \(\lim_{x\to\infty}(f(x)-x)=\lim_{x\to\infty}\frac{5}{\sqrt{x^2+5}+x}=0\), so the asymptote is \(y=x\). For the left-hand asymptote, \(\lim_{x\to-\infty}(f(x)+x)=\lim_{x\to-\infty}\frac{5}{\sqrt{x^2+5}-x}=0\), so the asymptote is \(y=-x\). Their slopes are the derivative limits from part c).

a) Local minimum at \((0,\sqrt{5})\). b) Since \(f''(x)=\frac{5}{(x^2+5)^{3/2}}>0\) for all real \(x\), there are no inflection points. c) \(\lim_{x\to\infty}f'(x)=1\) and \(\lim_{x\to-\infty}f'(x)=-1\). d) The slant asymptotes are \(y=x\) as \(x\to\infty\) and \(y=-x\) as \(x\to-\infty\). Their slopes are \(1\) and \(-1\), respectively.

- 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 \(x\to\infty\).

- 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 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

- Choose a denominator that is never zero. - Start with the slant-asymptote expression and add a term that approaches zero. - Use the y-intercept to determine the numerator of the remainder term. - The expression \(x^2+1\) is positive for every real \(x\).

1. To make the function defined for every real number, use a denominator with no real zeros, such as \(x^2+1\). 2. Write \(g(x)=-x+1+\frac{a}{x^2+1}\). The rational term approaches \(0\) as \(|x|\to\infty\), so the slant asymptote is \(y=-x+1\). 3. Use the y-intercept: \(2=g(0)=1+a\), so \(a=1\). 4. Therefore, \(g(x)=-x+1+\frac{1}{x^2+1}\). As one fraction, \(g(x)=\frac{-x^3+x^2-x+2}{x^2+1}\).

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

- At which value is the rational term undefined? - What happens to \(\frac{2}{x}\) as \(\lvert x\rvert\) becomes large? - The remaining linear term determines the slant asymptote.

1. The term \(\frac{2}{x}\) is undefined at \(x=0\), and its magnitude becomes unbounded as \(x\to 0\). Therefore, the vertical asymptote is \(x=0\). 2. As \(x\to\pm\infty\), \(\frac{2}{x}\to 0\). Therefore, \(f(x)\) approaches \(0.5x\), so the slant asymptote is \(y=0.5x\).

Vertical asymptote: \(x=0\); slant asymptote: \(y=0.5x\)

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

- Find the slope and intercept of the dashed line. - Use the vertical asymptote to choose the denominator of the remainder term. - Substitute a clear point such as \((1, 3)\) to determine the remaining constant.

1. The dashed line passes through the origin with slope \(1\), so the slant asymptote is \(y=x\). 2. The graph has a vertical asymptote at \(x=0\), so use \(f(x)=x+\frac{k}{x}\). 3. The graph passes through \((1, 3)\). Substituting gives \(3=1+k\), so \(k=2\). 4. Therefore, one possible function is \(f(x)=x+\frac{2}{x}\).

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

- 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 the sign of the given second derivative to classify the critical point. - Remember that \(\sqrt{x^2}=|x|\) when evaluating limits at infinity. - Complete the square inside the radical to identify the intercepts of the slant asymptotes.

1. The first derivative is \(f^{\prime}(x)=\frac{4x+2}{\sqrt{4x^2+4x+10}}\). Setting the numerator equal to zero gives \(x=-\frac{1}{2}\). Since \(f(-\frac{1}{2})=3\) and \(f^{\prime\prime}(x)>0\), the graph has a local and absolute minimum at \((-\frac{1}{2}, 3)\). 2. The expression \(4x^2+4x+10\) is positive for every real \(x\), so \(f^{\prime\prime}(x)>0\) everywhere. The concavity never changes, so there are no inflection points. 3. Divide numerator and denominator of \(f^{\prime}\) by \(|x|\). For \(x\to\infty\), the limit is \(2\); for \(x\to-\infty\), the limit is \(-2\). 4. Complete the square: \(f(x)=\sqrt{(2x+1)^2+9}\). As \(x\to\infty\), \(f(x)-(2x+1)\to0\), so one asymptote is \(y=2x+1\). As \(x\to-\infty\), \(f(x)-(-2x-1)\to0\), so the other is \(y=-2x-1\).

1. Absolute minimum at \((-\frac{1}{2}, 3)\) 2. \(f^{\prime\prime}(x)>0\) for all \(x\), so there are no inflection points. 3. \(\lim_{x\to\infty}f^{\prime}(x)=2\) and \(\lim_{x\to-\infty}f^{\prime}(x)=-2\) 4. \(y=2x+1\) and \(y=-2x-1\)

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

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

- 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: \(y=\frac{1}{3}x-2\)