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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">
Hey, Cynthia I found a calculus problem on locating the tangent line equation from our practice exam that will be on our upcoming exam, and I will explain it to you in detail if you are still having problems with the concept? 

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<div><div class="alert blue">
Sure, you know I need help with this section because I just do not have the time for a tutor or the drop-in lab.


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Question #1 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">

How do I use the definition of the derivative in helping with locating the tangent line? 

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

To use the definition of the derivative you have to follow this step:

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:
$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim_{h \to 0}\frac{[5(x+h)^2+10(x+h)-20]-[5x^2+10x+20]}{h}$
$f'(x)=\lim_{h \to 0}\frac{[5(x^2+2xh+h^2)-10(x+h)+20]-(5x^2+10x+20)}{h}$
$f'(x)=\lim_{h \to 0}\frac{(5x^2+10xh+5h^2-10x-10h+20-5x^2+10x-20)}{h}$
$f'(x)=\lim_{h \to 0}\frac{(10xh+5h^2-10h)}{h}$
$f'(x)=\lim_{h \to 0}\frac{h(10x+5h-10)}{h}$
$f'(x)={10x+5h-10}$
$f'(x)=10x+5(0)-10$
$f'(x)=10x-10$

Now that we have the derivative of the original formula, next we have to find the tangent line for the function 5x^2-10x+20.  First we have to evaluate the original function at the point x=5 for our b component of our linear formula as follows;

$b=5x^2-10x+20$
 $b=(5(5)^2+10(5)+20)$
 $b=95$
Next, we have to evaluate the derivative at the point x=5 for our slope or m-value to our linear formula.

$m=10x-10$
$m=10(5)-10$
$m=40$
Plugging your values into the formula of y=mx+b is your final step in creating your linear formula and now you can plug in any value near the point x=5 for an approximation:
y=40(x-5)+95

The formula above could be used to estimate a linear approximation for any give point close to x=5, so Cynthia you would chose a number and plug it into the x-value of the linear formula.
 
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<div><div class="alert blue">

Okay, thank you for simplifying for me how to find a tangent line when given a function and a value for x and on how to derive the formula.  from the function at any point.



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

No problem and we will talk after class to discuss our exam results.

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Okay, and thanks for everything.

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