Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<style>
body {
background-color: #eeeeee;
}

h1 {
color: maroon;
margin-left: 40px;
}

.gray {
margin-left: 50px ;
margin-right: 29%;
font-weight: 500;
color: #000000;
background-color: #cccccc;
border-color: #aaaaaa;
}

.blue {
display: inline-block;
margin-left: 29% ;
margin-right: 0%;
width: -webkit-calc(70% - 50px);
width: -moz-calc(70% - 50px);
width: calc(70% - 50px);
font-weight: 500;
color: #fff;
border-color: #336699;
background-color: #337799;
}

.left {
content:url("https://i.imgur.com/rUsxo7j.png");
width:50px;
border-radius: 50%;
float:left;
}
.right{
content:url("https://i.imgur.com/5ALcyl3.png"); width:50px;
border-radius: 50%;
display: inline-block;
vertical-align:top;
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</style>

<div id="container" style=" padding: 6px;
color: #fff;
border-color: #336699;
background-color: #337799;
display: flex;
justify-content: space-between;
margin-bottom:3px;">
<div>
<i class="fa fa-envelope fa-2x"></i>
</div>
<div>
<i class="fa fa-camera fa-2x"></i>
</div>

<div>
<i class="fa fa-comments fa-2x"></i>
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<div>
<i class="fa fa-address-card fa-2x" aria-hidden="true"></i>
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<i class="fa fa-phone fa-2x" aria-hidden="true"></i>
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<i class="fa fa-list-ul fa-2x" aria-hidden="true"></i>
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<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
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<div><img class="left"/><div class="alert gray">

I am studying for the Exam and I am having trouble finding the equation for the tangent line to the graph of a function at a point, and find the formula for the derivative of a function.

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<div><div class="alert blue">
Let's go over an example, I will take you set by step. Let me know if you have any questions.

An example of this would be
Consider the function f(x) = $4x^2−8x$.
(a) Find the equation for the tangent line to the graph of this function at x = 2. Use only the definition of
the derivative.
(b) Find the formula for the derivative of this function at any point. Use only the definition of the derivative.
</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

The limit definition is f'(x)= lim h➙0 $\frac{f(a+h)-f(a)}{h}$. Correct?

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I understand the first step would be finding the function but this is where I get lost.

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<div><div class="alert blue">

I would recommend to find f'(x) first so you have all the tools to find the tangent of line.
The first step is to plug in the f'(x) into f(x) to find the formula.
1) f(x)=h➙0 $4x^2−8x$ so
f'(x)= h➙0 $\frac{4(x+h)^2−8(x+h)}{h}$
2) You begin to distrubute all of the numbers so we can try to cancel out numbers.
= h➙0 $\frac{[4x^2+8hx-4h^2-8x-8h]}{h}$

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<div><img class="left"/><div class="alert gray">

I see that when I was praticing that I made a mistake on the factoring portion.

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<div><div class="alert blue">

Yes be careful when you are factoring out the problems.

The next steps would be:
3) = h➙0 $\frac{[4x^2+8hx-4h^2-8x-8h]}{h}$ so would then cancel out the ones that are the same.
4) = h➙0 $\frac{[4x^2+8hx-4x^2-8x]}{h}$
5) = h➙0 $\frac{[8hx-8h]}{h}$
6) = h➙0 $\frac{[h(8x-8]}{h}$ (the h's cancel)
7) f'(x)= 8x-8

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<div><img class="left"/><div class="alert gray">

Okay so that I have find the tangent line. What are the steps?

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<div><div class="alert blue">

The steps for the tangent line is you would have to take the equation from previous problem and plug it in.

L(x)= f(a)+f'(a)(x-a)
L(x)= f(2)+f'(2)(x-2)
= $4(2)^2−8(2)$+(2-2)
= 0
The tangent of the line is 0.

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