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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
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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) I found the domian by setting the denominator equal to zero. 

$x^3=0$ so x=o. This means that x=0 the function is undefined. 

Domain[x<0 or x>0 ]

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(2) Find all $x$- and $y$-intercepts. 
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(2) There is no y intercept because the function will be undefined when you set x=0. 

The x intercept is found when you set the function equal to zero. 

$0=\frac{12x^2-16}{x^3}$

I set the numerator equal to zero, 
$12x^2-16=0$
$12x^2=16$
$x^2=16/12$
$x=-√(4/3)$ and $x=√(4/3)$ are the x intercepts.





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(3) Find all equations of horizontal asymptotes. 
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(3) The equation of the horizontal asymptote is y=0
I found this because the horizontal asymptote is y=0 is when the numerators degree is one less than the degree of the numerator. 


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(4) Find all equations of vertical asymptotes. 
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(4) The equation of the verticle asymptote is x=0, because this is when the function is undefined.

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(5) Find the interval(s) where $f$ is increasing.
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(5) I found the intervals by finding the critical values of the function by settinf f'(x)= 0 or undefined

f is increasing from -2<x<0 and 0<x<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 local maxima is (2,4)
I got this by inputting the extreme point back into the function. 

$y=12(2)^2-16/(2)^3$ 

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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 x values is -2, I got this by utilizing the other extreme point. the coordinate is (-2,-4)

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(8) Find the interval(s) on which the graph is concave downward. 
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(8) I first found the inflection points then plugged in to see if it was positive or negative. The graph is concave down in the interval (-infinty<x< -2√2) and (0<x<2√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) I found the inflection points by making the f"=0
and finding where f" is undefined by making the denominator equal to zero. 

The inflection points are at x=-2√2 and x=2√2, 
Then I plugged those into to find the y coordinate. 

Points are (-2√2, -5/√2) and (2√2, 5/√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/ggplofn0i9?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>






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