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title: 'Math 181 Miniproject 7: The Shape of a Graph'

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Math 181 Miniproject 7: The Shape of a Graph.md
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tags: MATH 181
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Math 181 Miniproject 7: The Shape of a Graph
===

**Overview:**   In this miniproject you will be using the techniques of calculus to find the behavior of a graph. 

**Prerequisites:** The project draws heavily from the ideas of Chapter 1 and $2.8$ together with ideas and techniques of the first and second derivative tests from $3.1$. 

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We are given the functions 
$$
f(x)=\frac{12x^2-16}{x^3},\qquad f'(x)=-\frac{12(x^2-4)}{x^4},\qquad f''(x)=\frac{24(x^2-8)}{x^5}.
$$ 
The questions below are about the function $f(x)$. Answer parts (1) through (10) below. If the requested feature is missing, then explain why. Be sure to include the work/test that you used to rigorously reach your conclusion. It is not sufficient to refer to the graph.
(1) State the function's domain. 
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(1)
$(-infinity,0)U(0,infinity)$

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(2) Find all $x$- and $y$-intercepts. 
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(2)
$0=\frac{12x^{2}-16}{x^{3}}$
$x=\sqrt{\frac{4}{3}}$
$x$-intercept $(\frac{2}{\sqrt{3}},0)$
$f\left(0\right)=\frac{12\left(0\right)^{2}-16}{\left(0\right)^{3}}$
$f\left(0\right)=$undefined
$f(x)$ has a vertical asymptote at $x=0$


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(3) Find all equations of horizontal asymptotes. 
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(3)
$f(x)=\frac{12x^2-16}{x^3}$
$\frac{x^{2}}{x^{3}}$
$f(x)$ has a horizontal asymptote at $y=0$
$f'(x)=-\frac{12(x^2-4)}{x^4}$
$\frac{x^{2}}{x^{4}}$
$f'(x)$ has a horizontal asymptote at $y=0$
$f''(x)=\frac{24(x^2-8)}{x^5}$
$\frac{x^{2}}{x^{5}}$
$f''![](https://i.imgur.com/o0FS0e9.png)
![](https://i.imgur.com/a30Mah1.png)
(x)$ has a horizontal asymptote at $y=0$


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(4) Find all equations of vertical asymptotes. 
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(4)
$f(x)=\frac{12x^2-16}{x^3}$
$0=x^{3}$
$f(x)$ has a vertical asymptote at $x=0$
$f'(x)=-\frac{12(x^2-4)}{x^4}$
$0=x^{4}$
$f'(x)$ has a vertical asymptote at $x=0$
$f''(x)=\frac{24(x^2-8)}{x^5}$
$0=x^{5}$
$f''(x)$ has a vertical asymptote at $x=0$

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(5) Find the interval(s) where $f$ is increasing.
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(5)
$f'(x)=-\frac{12(x^{2}-4)}{x^{4}}$
$0=-\frac{12(x^{2}-4)}{x^{4}}$
Note $x$ is undefined at $0$.
Critical values are $x=-2,2$
After conducting the first derivative test I determined the intervals where $f$ is increasing are $[-2,0)$ ]and $(0,2]$.

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(6) Find the $x$-value(s) of all local maxima. (Find exact values, and not decimal representations) 
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(6)
$f'(x)=-\frac{12(x^{2}-4)}{x^{4}}$
$0=-\frac{12(x^{2}-4)}{x^{4}}$
Note $x$ is undefined at $0$.
Critical values are $x=-2,2$
$f(x)$ is increasing from $(0,2]$ and decreasing from $[2,infinity)$. Therefore, 2 is the only local maxima.

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(7) Find the $x$-value(s) of all local minima. (Find exact values, and not decimal representations)
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(7)
$f'(x)=-\frac{12(x^{2}-4)}{x^{4}}$
$0=-\frac{12(x^{2}-4)}{x^{4}}$
Note $x$ is undefined at $0$.
Critical values are $x=-2,2$
$f(x)$ is decreasing from $(-infinity,-2]$ and increasing from $[-2,und.)$. Therefore, -2 is the only local minima.

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(8) Find the interval(s) on which the graph is concave downward. 
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(8)
$f''(x)=\frac{24(x^2-8)}{x^5}$
$0=\frac{24(x^{2}-8)}{x^{5}}$
Note $x$ is undefined at $0$.
Critical values are $x=+\sqrt{8},-\sqrt{8}$
The graph is concave downward on the intervals $\left(-\infty,-\sqrt{8}\right]$ and $\left(0,\sqrt{8}\right]$.


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(9) State the $x$-value(s) of all inflection points. (Find exact values, and not decimal representations)
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(9)
$f''(x)=\frac{24(x^2-8)}{x^5}$
$0=\frac{24(x^{2}-8)}{x^{5}}$
Note $x$ is undefined at $0$.
Critical values are $x=+\sqrt{8},-\sqrt{8}$
The inflection points of the graph are the critical values of the graph:$x=+\sqrt{8},-\sqrt{8}$

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(10) Include a sketch of the graph of $y=f(x)$.  Plot the different segments of the graph using the color code below.
* **blue:** $f'>0$ and $f''>0$
* **red:** $f'<0$ and $f''>0$
* **black:** $f'>0$ and $f''<0$
* **gold:** $f'<0$ and $f''<0$

(In Desmos you could restrict the plot $y=f(x)$ on the interval $[2,3]$ by typing $y=f(x)\{2\le x\le 3\}$.) Be sure to set the bounds on the graph so that the features of the graph that you listed above are easy to see.
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(10)
![](https://i.imgur.com/QJZSCVu.png)





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