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Consider the function $f\left(x\right)=5-{x}^{2\text{/}3}.$ Determine the point on the graph where a cusp is located. Determine the end behavior of $f.$
The function $f$ has a cusp at $\left(0,5\right)$ $\underset{x\to {0}^{-}}{\text{lim}}{f}^{\prime}\left(x\right)=\infty ,$ $\underset{x\to {0}^{+}}{\text{lim}}{f}^{\prime}\left(x\right)=\text{\u2212}\infty .$ For end behavior, $\underset{x\to \text{\xb1}\infty}{\text{lim}}f\left(x\right)=\text{\u2212}\infty .$
For the following exercises, examine the graphs. Identify where the vertical asymptotes are located.
For the following functions $f\left(x\right),$ determine whether there is an asymptote at $x=a.$ Justify your answer without graphing on a calculator.
$f\left(x\right)=\frac{x+1}{{x}^{2}+5x+4},a=\mathrm{-1}$
$f\left(x\right)=\frac{x}{x-2},a=2$
Yes, there is a vertical asymptote
$f\left(x\right)={\left(x+2\right)}^{3\text{/}2},a=\mathrm{-2}$
$f\left(x\right)={\left(x-1\right)}^{\mathrm{-1}\text{/}3},a=1$
Yes, there is vertical asymptote
$f\left(x\right)=1+{x}^{\mathrm{-2}\text{/}5},a=1$
For the following exercises, evaluate the limit.
$\underset{x\to \infty}{\text{lim}}\frac{2x-5}{4x}$
$\underset{x\to \infty}{\text{lim}}\frac{{x}^{2}-2x+5}{x+2}$
$\infty $
$\underset{x\to \text{\u2212}\infty}{\text{lim}}\frac{3{x}^{3}-2x}{{x}^{2}+2x+8}$
$\underset{x\to \text{\u2212}\infty}{\text{lim}}\frac{{x}^{4}-4{x}^{3}+1}{2-2{x}^{2}-7{x}^{4}}$
$-\frac{1}{7}$
$\underset{x\to \infty}{\text{lim}}\frac{3x}{\sqrt{{x}^{2}+1}}$
$\underset{x\to \text{\u2212}\infty}{\text{lim}}\frac{\sqrt{4{x}^{2}-1}}{x+2}$
$\mathrm{-2}$
$\underset{x\to \infty}{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}$
$\underset{x\to \text{\u2212}\infty}{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}$
$\mathrm{-4}$
$\underset{x\to \infty}{\text{lim}}\frac{2\sqrt{x}}{x-\sqrt{x}+1}$
For the following exercises, find the horizontal and vertical asymptotes.
$f\left(x\right)=x-\frac{9}{x}$
Horizontal: none, vertical: $x=0$
$f\left(x\right)=\frac{1}{1-{x}^{2}}$
$f\left(x\right)=\frac{{x}^{3}}{4-{x}^{2}}$
Horizontal: none, vertical: $x=\text{\xb1}2$
$f\left(x\right)=\frac{{x}^{2}+3}{{x}^{2}+1}$
$f\left(x\right)=\text{sin}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right)$
Horizontal: none, vertical: none
$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\left(3x\right)+\text{cos}\left(5x\right)$
$f\left(x\right)=\frac{x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(x\right)}{{x}^{2}-1}$
Horizontal: $y=0,$ vertical: $x=\text{\xb1}1$
$f\left(x\right)=\frac{x}{\text{sin}\left(x\right)}$
$f\left(x\right)=\frac{1}{{x}^{3}+{x}^{2}}$
Horizontal: $y=0,$ vertical: $x=0$ and $x=\mathrm{-1}$
$f\left(x\right)=\frac{1}{x-1}-2x$
$f\left(x\right)=\frac{{x}^{3}+1}{{x}^{3}-1}$
Horizontal: $y=1,$ vertical: $x=1$
$f\left(x\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x}{\text{sin}\phantom{\rule{0.1em}{0ex}}x-\text{cos}\phantom{\rule{0.1em}{0ex}}x}$
$f\left(x\right)=x-\text{sin}\phantom{\rule{0.1em}{0ex}}x$
Horizontal: none, vertical: none
$f\left(x\right)=\frac{1}{x}-\sqrt{x}$
For the following exercises, construct a function $f\left(x\right)$ that has the given asymptotes.
$x=1$ and $y=2$
Answers will vary, for example: $y=\frac{2x}{x-1}$
$x=1$ and $y=0$
$y=4,$ $x=\mathrm{-1}$
Answers will vary, for example: $y=\frac{4x}{x+1}$
For the following exercises, graph the function on a graphing calculator on the window $x=\left[\mathrm{-5},5\right]$ and estimate the horizontal asymptote or limit. Then, calculate the actual horizontal asymptote or limit.
[T] $f\left(x\right)=\frac{x+1}{{x}^{2}+7x+6}$
[T] $\underset{x\to \text{\u2212}\infty}{\text{lim}}{x}^{2}+10x+25$
$\infty $
[T] $\underset{x\to \text{\u2212}\infty}{\text{lim}}\frac{x+2}{{x}^{2}+7x+6}$
[T] $\underset{x\to \infty}{\text{lim}}\frac{3x+2}{x+5}$
$y=3$
For the following exercises, draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior.
$y=3{x}^{2}+2x+4$
$y=\frac{2x+1}{{x}^{2}+6x+5}$
$y=\frac{{x}^{2}+x-2}{{x}^{2}-3x-4}$
$y=2x\sqrt{16-{x}^{2}}$
$y=\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x},$ on $x=\left[\mathrm{-2}\pi ,2\pi \right]$
$y={e}^{x}-{x}^{3}$
$y=x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}x,x=\left[\text{\u2212}\pi ,\pi \right]$
$y=x\phantom{\rule{0.1em}{0ex}}\text{ln}\left(x\right),x>0$
$y={x}^{2}\text{sin}\left(x\right),x=\left[\mathrm{-2}\pi ,2\pi \right]$
For $f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$ to have an asymptote at $y=2$ then the polynomials $P\left(x\right)$ and $Q\left(x\right)$ must have what relation?
For $f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}$ to have an asymptote at $x=0,$ then the polynomials $P\left(x\right)$ and $Q\left(x\right).$ must have what relation?
$Q\left(x\right).$ must have have ${x}^{k+1}$ as a factor, where $P\left(x\right)$ has ${x}^{k}$ as a factor.
If ${f}^{\prime}\left(x\right)$ has asymptotes at $y=3$ and $x=1,$ then $f\left(x\right)$ has what asymptotes?
Both $f\left(x\right)=\frac{1}{\left(x-1\right)}$ and $g\left(x\right)=\frac{1}{{\left(x-1\right)}^{2}}$ have asymptotes at $x=1$ and $y=0.$ What is the most obvious difference between these two functions?
$\underset{x\to {1}^{-}}{\text{lim}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\underset{x\to {1}^{-}}{\text{lim}}g\left(x\right)$
True or false: Every ratio of polynomials has vertical asymptotes.
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