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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<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?

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<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?

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<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.
$f'(x)=10x+5(0)+30$
$f'(x)=10x+30$








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<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.

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

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<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.

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<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.







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