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)
$x^3=0$
$x=0$
Domain: $x=(-∞, 0), (0, ∞)$

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(2) Find all $x$- and $y$-intercepts.
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(2)
$0=\frac{12x^2-16}{x^3}$
$0=12x^2-16$
$0=4(3x^2-4)$
$3x^2=4$
$x=-\sqrt(\frac{4}{3}), \sqrt(\frac{4}{3})$
x-intercepts: $(-\sqrt(\frac{4}{3}), 0), (-\sqrt(\frac{4}{3}), 0)$

$y=\frac{12(0)^2-16}{(0)^3}$
y is undefined at x=0, so there is no y-intercept

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(3) Find all equations of horizontal asymptotes.
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(3)
Since the degree of the denomiinator is greater than the degree of the numerator, the horizontal asymptote of $f$ is $y=0$

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(4) Find all equations of vertical asymptotes.
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(4)
$x^3=0$
$x=0$
vertical asymptote: $x=0$

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(5) Find the interval(s) where $f$ is increasing.
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(5)
$f'$ is undefined at $x=0$
$-\frac{12(x^2-4)}{x^4}=0$
$x^2-4=0$
$x=-2, 2$
$f$ is increasing on the intervals $(-2, 0)$ 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)
$x=2$ is the local maxima as seen in the previous image.
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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)
$x=-2$ is the local minima as seen in the previous image.
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(8) Find the interval(s) on which the graph is concave downward.
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(8)
The function $f''$ is undefined at $x=0$
$\frac{24(x^2-8)}{x^5}=0$
$(x^2-8)=0$
$x=-2\sqrt2, 2\sqrt2$
$f$ is concave downward on the intervals $(-\infty, -2\sqrt2)$ and $(0, 2\sqrt2)$

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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 are $x= -2\sqrt2, 0, 2\sqrt2$ as seen in the previous image.

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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/hflv45o6t8?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

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