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
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**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)

$f(0)=\frac{12(0)^2-16}{(0)^3}$
$f(0)=\frac{-16}{0}$ =undefined, which means
$x\neq0$
$(-\infty,0],[0,\infty)$
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(2) Find all $x$- and $y$-intercepts.
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(2)
To find the x intercepts, just set the numerator of the function to 0.
$12x^2-16=0$
$12x^2=16$
$x^2=\frac{16}{12}$
$\sqrt{x^2}=\frac{\sqrt{16}}{\sqrt{12}}$
$x=\pm\frac{4}{\sqrt{12}}$
Simplified to $x=\pm\frac{2}{\sqrt{3}}$
To find the y intercepts, just plug in 0 for all the x values
$f(0)=\frac{12(0)^2-16}{(0)^3}$
$f(0)=\frac{-16}{0}$, which is undefined, so there is no y intercepts. This makes since because in the functions domian, it tells us that x cannot equal 0.
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(3) Find all equations of horizontal asymptotes.
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(3)
In order to do this, you need to take the degree of the numerator (=2) and the degree of the denominator (=3). Then $2<3$
The degree of the numerator is less than the degree of the denominator. Therefore, the horizontal asymptote of the equation is $y=0$
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(4) Find all equations of vertical asymptotes.
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(4)
To find the vertical asymptote, set the denominator to equal 0
$x^3=0$
This makes the vertical asymptote of the equation $x=0$
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(5) Find the interval(s) where $f$ is increasing.
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(5)
By setting the derivative to 0 you get
$12=0$ No Solution
$x^2-4=0$
$x=\pm2$, these are the critical values, but we also know that the denominator shows a vertical asymptote at 0, so when we put it values on the 1st derivative test, we need to include 0 as well.
By using the first derivative test, you are able to find that f is increasing at the intervals
$[-2,0],[0,2]$

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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)
Based on our 1st derivative test, we know that the funciton has no local maxima when $x=2$
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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)
Based on our first derivative test, we know that the funtion has a local minima when $x=-2$
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(8) Find the interval(s) on which the graph is concave downward.
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(8)
By doing the 2nd derivative test we know that there will be critical values at $\pm2\sqrt{2}$ and there is still a vertical asymptote at 0, so the 0 will be present in the second derivative test
From plugging in test numbers, we found that the intervals where the graph is concave downward at
$(-\infty,-2\sqrt{2}],[0,2\sqrt{2}]$
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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)
The inflection points found by doing the second derivative test show up when the concave changes, which is at
$x=-2\sqrt{2},2\sqrt{2}$
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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)
<iframe src="https://www.desmos.com/calculator/9qzhx4c3jt?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

Based on this graph, all of the information above seems to be correct based on the derivative tests.

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