# 11.3: Exercises

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## Exercise $$\PageIndex{1}$$

Find the domain, the vertical asymptotes and removable discontinuities of the functions.

1. $$f(x)=\dfrac{2}{x-2}$$
2. $$f(x)=\dfrac{x^2+2}{x^2-6x+8}$$
3. $$f(x)=\dfrac{3x+6}{x^3-4x}$$
4. $$f(x)=\dfrac{(x-2)(x+3)(x+4)}{(x-2)^2(x+3)(x-5)}$$
5. $$f(x)=\dfrac{x-1}{x^3-1}$$
6. $$f(x)=\dfrac{2}{x^3-2x^2-x+2}$$
1. domain $$D = \mathbb{R} − \{2\}$$, vertical asymptote at $$x = 2$$, no removable discontinuities
2. $$D = \mathbb{R}−\{2, 4\}$$, vertical asympt. at $$x = 2$$ and $$x = 4$$, no removable discont.
3. $$D = \mathbb{R}− \{−2, 0, 2\}$$, vertical asympt. at $$x = 0$$ and $$x = 2$$, removable discont. at $$x = −2$$
4. $$D = \mathbb{R} − \{−3, 2, 5\}$$, vertical asympt. at $$x = 2$$ and $$x = 5$$, removable discont. at $$x = −3$$
5. $$D = \mathbb{R} − \{1\}$$, no vertical asympt., removable discont. at $$x = 1$$
6. $$D = \mathbb{R} − \{−1, 1, 2\}$$, vertical asympt. at $$x = −1$$ and $$x = 1$$ and $$x = 2$$, no removable discont.

## Exercise $$\PageIndex{2}$$

Find the horizontal asymptotes of the functions.

1. $$f(x)=\dfrac{8x^2+2x+1}{2x^2+3x-2}$$
2. $$f(x)=\dfrac{1}{(x-3)^2}$$
3. $$f(x)=\dfrac{x^2+3x+2}{x-1}$$
4. $$f(x)=\dfrac{12x^3-4x+2}{-3x^3+2x^2+1}$$
1. $$y = 4$$
2. $$y = 0$$
3. no horizontal asymptote (asymptotic behavior $$y = x + 4$$)
4. $$y = −4$$

## Exercise $$\PageIndex{3}$$

Find the $$x$$- and $$y$$-intercepts of the functions.

1. $$f(x)=\dfrac{x-3}{x-1}$$
2. $$f(x)=\dfrac{x^3-4x}{x^2-8x+15}$$
3. $$f(x)=\dfrac{(x-3)(x-1)(x+4)}{(x-2)(x-5)}$$
4. $$f(x)=\dfrac{x^2+5x+6}{x^2+2x}$$
1. $$x$$-intercept at $$x = 3$$, $$y$$-intercept at $$y = 3$$
2. $$x$$-intercepts at $$x = 0$$ and $$x = −2$$ and $$x = 2$$, $$y$$-intercept at $$y = 0$$
3. $$x$$-intercepts at $$x = −4$$ and $$x = 1$$ and $$x = 3$$, $$y$$-intercept at $$y = \dfrac{6}{5}$$
4. $$x$$-intercept at $$x = −3$$ (but not at $$x = −2$$ since $$f(−2)$$ is undefined), no $$y$$-intercept since $$f(0)$$ is undefined

## Exercise $$\PageIndex{4}$$

Sketch the graph of the function $$f$$ by using the domain of $$f$$, the horizontal and vertical asymptotes, the removable singularities, the $$x$$- and $$y$$-intercepts of the function, together with a sketch of the graph obtained from the calculator.

1. $$f(x)=\dfrac{6x-2}{2x+4}$$
2. $$f(x)=\dfrac{x-3}{x^3-3x^2-6x+8}$$
3. $$f(x)=\dfrac{x^4-10x^2+9}{x^2-3x+2}$$
4. $$f(x)=\dfrac{x^3-3x^2-x+3}{x^3-2x^2}$$
1. $$D = \mathbb{R} − \{2\}$$, horizontal asympt. $$y = 3$$, vertical asympt. $$x = −2$$, no removable discont., $$x$$-intercept at $$x = \dfrac 1 3$$, $$y$$-intercept at $$y = \dfrac{-1}{2}$$, graph:

1. $$f(x)=\dfrac{x-3}{(x-4)(x-1)(x+2)}$$ has domain $$D = \mathbb{R} − \{−2, 1, 4\}$$, horizontal asympt. $$y = 0$$, vertical asympt. $$x = −2$$ and $$x = 1$$ and $$x = 4$$, no removable discont., $$x$$-intercept at $$x = 3$$, $$y$$-intercept at $$y = \dfrac{-3}{8} = −0.375$$, graph:

1. $$f(x)=\dfrac{(x-3)(x+3)(x-1)(x+1)}{(x-2)(x-1)}$$ has domain $$D = \mathbb{R} − \{1, 2\}$$, no horizontal asympt., vertical asympt. $$x = 2$$, removable discont. at $$x = 1$$, $$x$$-intercept at $$x = −3$$ and $$x = −1$$ and $$x = 3$$, $$y$$-intercept at $$y = \dfrac 9 2 = 4.5$$, graph:

1. $$f(x)=\dfrac{(x-3)(x-1)(x+1)}{x^{2}(x-2)}$$ has domain $$D = \mathbb{R} − \{0, 2\}$$, horizontal asympt. $$y = 1$$, vertical asympt. $$x = 0$$ and $$x = 2$$, no removable discont., $$x$$-intercepts at $$x = −1$$ and $$x = 1$$ and $$x = 3$$, no $$y$$-intercept since $$f(0)$$ is undefined, graph:

Note that the graph intersects the horizontal asymptote $$y = 1$$ at approximately $$x \approx-2.3$$ and approaches the asymptote from above

## Exercise $$\PageIndex{5}$$

Find a rational function $$f$$ that satisfies all the given properties.

1. vertical asymptote at $$x=4$$ and horizontal asymptote $$y=0$$
2. vertical asymptotes at $$x=2$$ and $$x=3$$ and horizontal asymptote $$y=5$$
3. removable singularity at $$x=1$$ and no horizontal asymptote
1. for example $$f(x)=\dfrac{1}{x-4}$$
2. for example $$f(x)=\dfrac{5 x^{2}}{x^{2}-5 x+6}$$
3. for example $$f(x)=\dfrac{x^{2}-x}{x-1}$$