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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) Because x=0, the given function is undefined so domain is (-infinity,0)U(0,infinity)

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(2) Find all $x$- and $y$-intercepts. 
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(2) for f(x)=0
12x^2- (16/x^3) =0
x^2=16/12 = 4/3
x= 2/sqrt3
x intercepts are (2/sqrt3,0) and (-2/sqrt3,0)
for our y intercepts we plug x=0 but f(x) is undefined at x =0 so y intercept DNE


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(3) Find all equations of horizontal asymptotes. 
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(3)y=L is a horizontal asymptote of the function y=f(x)
as the lim approaches infinity (12/x - 16/x^3)=0
and also as the limit approaches negative infinity it equals 0
which means the horizontal asymptote is y=0

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(4) Find all equations of vertical asymptotes. 
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(4)x=L is a vertical asymptote of the function y=f(x) if the limit at this point is infinite
as the limit approaches to the right of 0, (12/x-16/x^3)= -infinity
since the limit is infinite, x=0 is the vertical asymptote

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(5) Find the interval(s) where $f$ is increasing.
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(5) f''= 12 (2-x)(x+2)/x^2
x= (-2,0)U(o,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) The maxima is 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) The minima is x=-2

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(8) Find the interval(s) on which the graph is concave downward. 
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(8)f''(x)=24(x^2 -8)/x^5
(-infinity,-2sqrt2),(0,2sqrt2)

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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 the y values
-2sqrt2, 2sqrt2

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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/abSdM9s.jpg)







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