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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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:::info
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. 
:::

The domain for $f$ is all real numbers except for $0$. The function found in the numerator is a polynomial, polynomials are able to use all real numbers as the domain. Dividing by $0$ is not allowed, hence the domain for $f$ is all real numbers except for $0$.



:::info
(2) Find all $x$- and $y$-intercepts. 
:::

**$x$-intercepts:**
$f(x)=\frac{12x^2-16}{x^3}$

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

$0=12𝑥^2−16$
$16=12x^2$
$1.333333333=𝑥^2$
$x=±1.154700538$

**$y$-intercepts:**
$f(x)=\frac{12x^2-16}{x^3}$

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

There are no $y$-intercepts because we are not allowed to divide by 0.


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(3) Find all equations of horizontal asymptotes. 
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There is a horizontal asymptote at $y=0$.
According to the rules of degrees of rational functions, if the degree of the numerator is larger or greater than the denominator then the horizontal asymptote would be equal to 0.

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

$0=x^3$
$\sqrt[3]{x^3}$=$\sqrt[3]{0}$
$x=0$

There is a vertical asymptote at x=0.

:::info
(5) Find the interval(s) where $f$ is increasing.
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**Using the first derivative test, find the critical values:**
$f'(x)=-\frac{12(x^2-4)}{x^4}$

$-12x(x^2-4)=0$
$x^2-4=0$
$x^2=4$
$\sqrt{x^2}=\sqrt{4}$
$x=±2$

**Now for $x^4$:**
$x^4=0$
$\sqrt[4]{x^4}=\sqrt[4]{0}$
$x=0$

The critical values are $-2, 0, 2$


Test the values before, after and between the critical values: 
$f′(−5)=−0.4032$ 
$f′(−1)=36$
$f′(1)=36$
$f′(5)=−0.4032$

$f$ is increasing between $x= -2$ and $0$ and also between $x=0$ and $2$.

![](https://i.imgur.com/8saWf63.png)


:::info
(6) Find the $x$-value(s) of all local maxima. (Find exact values, and not decimal representations) 
:::

There is a local maximum at $f (2)$
$f(x)=\frac{12x^2-16}{x^3}$

$f(2)=\frac{12(2)^2-16}{2^3}$

$f(2)=\frac{48-16}{8}$
$f(2)=4$

There is local maxima at (2, 4).

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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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There is a local minima at $f (-2)$
$f(x)=\frac{12x^2-16}{x^3}$

$f(-2)=\frac{12(-2)^2-16}{-2^3}$

$f(-2)=\frac{48-16}{-8}$
$f(-2)=-4$

There is local minima at (-2, -4).






:::info
(8) Find the interval(s) on which the graph is concave downward. 
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**Using the second derivative test:**

$f''(x)=\frac{24(x^2-8)}{x^5}$
$24(x^2-8)=0$
$𝑥^2−8=0$ 
$𝑥^2=8$ 
$\sqrt{𝑥2}=\sqrt{8}$ 
$x=±2\sqrt{2}$

**Now for the denominator:**

$x^5=0$
$\sqrt[5]{x^5}=\sqrt[5]{0}$
$x=0$

**To find out the concavity we plug in these values:**
$𝑓′′(−5)=−0.13056$
$𝑓′′(5)=0.13056$
$𝑓′′(−1)=168$
$𝑓′′(1)=−168$

The graph is concave down between $x=−∞$ and $x=−2\sqrt{2}$ as well as between $x=0$ and $2\sqrt{2}$

![](https://i.imgur.com/aNYzrS9.png)



:::info
(9) State the $x$-value(s) of all inflection points. (Find exact values, and not decimal representations)
:::

$f''(x)=\frac{24(x^2-8)}{x^5}$
$24(x^2-8)=0$
$𝑥^2−8=0$ 
$𝑥^2=8$ 
$\sqrt{𝑥2}=\sqrt{8}$ 
$x=±2\sqrt{2}$


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$x^5=0$
$\sqrt[5]{x^5}=\sqrt[5]{0}$
$x=0$

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**To find out the concavity we plug in these values:**
$𝑓′′(−5)=−0.13056$
$𝑓′′(5)=0.13056$
$𝑓′′(−1)=168$
$𝑓′′(1)=−168$

There are inflection points at $f(−2√2)$, $f(0)$, and $f(2√2)$. Moreover, the inflection points are: $(−2\sqrt{2},0)$ and $(2\sqrt{2},0)$. 
There cannot be one specific point at 0 since there is a vertical asymptote at $x=0$ which means it would be undefined.

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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/opPQf4Z.png)







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To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.



