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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)$(-∞,0)U(0,∞)

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(2) Find all $x$- and $y$-intercepts. 
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(2)
$f(x)=\frac{12x^2-4}{x^4}$
$12(x+2)(x-2)$
$x-2=0$
$x=2  (x-intercept)$
$x+2=0$
$x=-2 (x-intercept)$
$(x^4)=0$
y=0 (no y-intercepts)






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(3) Find all horizontal asymptotes.
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(3)
$f(x)=\frac{12x^2-16}{x^3}$ 
Applying the rules of horizontal asymptotes, the horizontal asymptotes is y=0.

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(4) Find all equations of vertical asymptotes. 
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(4)
$f(x)=\frac{12x^2-16}{x^3}$ 
Applying the rules of vertical asymptotes, the vertical asymptotes is x=0. 

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(5) Find the interval(s) where $f$ is increasing.
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(5)f is increasing on the interval from (-∞,-2) and increasing (2,∞).
![](https://i.imgur.com/EnTjqLs.jpg)


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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 local maxima is x=-2.
![](https://i.imgur.com/EnTjqLs.jpg)
Using the number line to determine where the local maximum is locating when the sign from f' switches from positive to negative.  Therefore, 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 local minimum is x=2.
![](https://i.imgur.com/EnTjqLs.jpg)
Using the number line to determine where the local maximum is locating when the sign from f' switches from negative to positive.  Therefore x=2.
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(8) Find the interval(s) on which the graph is concave downward. 
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(8)The graph concaves downward from $(-∞,-2\sqrt{2})$ and from $(0,2/sqrt{2}).
![](https://i.imgur.com/T9TegHb.jpg)




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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 when the concavity changes from positive to negative, therefore the inflection points are + and - 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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![](https://i.imgur.com/EQYm5KD.png)








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