Math 181 Miniproject 3: Texting Lesson.md
---
My lesson Topic
===

<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;
}
</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>
</div>
<div>
<i class="fa fa-address-card fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-phone fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-list-ul fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
</div>
</div>

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

Hello, Jessica did you understand how to find the equation for the tangent line to the graph of the function 5x^2+10x+20 at the point x=5, and find the formula for
the derivative of a function, using the limit-based definition of the derivative?

</div></div>

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

No, I did not understand how to find the tangent line to the graph for the function 5x^2-10x+20 at the point x=5, and I was hoping that you understood?

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">
Use the formula for determining the limit as h approaches zero, for the equation 5x^2+10x+20 for determining the tangent line which is:

$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim_{h \to 0}\frac{[5(x+h)^2+10(x+h)-20][5x^2+10x+20]}{h}$
$f'(x)=\lim_{h \to 0}\frac{5(x^2+2xh+h^2)+10(x+h)+20-(5x^2+10x+20)}{h}$
$f'(x)=\lim_{h \to 0}\frac{(5x^2+10xh+5h^2+10x+10h-5x^2-10x+20)}{h}$
$f'(x)=\lim_{h \to 0}\frac{(10xh+5h^2+10h+20)}{h}$
$f'(x)=\lim_{h \to 0}\frac{(10xh+5h+30)}{h}$
$f'(x)=\lim_{h \to 0}{10x+5h+30}$
At this point you take your limit as it approaches zero.
$f'(x)=10x+5(0)+30$
$f'(x)=10x+30$

</div></div>

<div><img class="left"/><div class="alert gray">
Next, you nee to use the tangent line approximation formula which is l(x)=f(a)+f'(a)(x-a) and plug in the values needed to find your tangent line equation.

</div></div>

<div><div class="alert blue">
Can you refresh my memory on how to plug in the appropiate values to find my tangent line?

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">
$l(x)=f(a)+f'(a)(x-a)$
$l(x)=f(5)+f'(5)(x-5)$
$l(x)=(5x^2+10x+20)+(10x-10)(x-5)$
And now you can plug-in any value near x=5 to find a linear approximation of that value.

</div></div>

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

Wow, you really simplified the processes for me. I am so glad that we are study buddies.

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

No problem and i'll see you in class.

</div></div>

---

To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.