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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 Fernando, I just realized that the Midterm is tomorrow and have no idea how to do number 2, would you mind helping me out?

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

Ok, number 2 has been reviewed by Dr.Ballif during class, so luckily it should not be anything new.

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

Okay good! Thank you so much.

</div></div> 

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

OK, So to get started, number 2 is a 2 parter.
According to question 2 we are supposed to first find the tangent line at $x=5$ for the function: $f\left(x\right)=3x^{2}-5x+10$
Althought it may seem counter-intuitive, I would first Solve part B. Then using Part B, we can solve for part A. 

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<div><img class="left"/><div class="alert gray">
Ok, So by the limit formula, do you mean the one that looks like this: 

$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

</div></div> 

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

Yes, that one. Now using the function given to us, we will be plugging it in to the original limit formula. 
So once you plug in the function, it should look like this: 

$\lim_{h \to 0}\frac{[3(x+h)^2+5(x+h)+(10)]-[3(x)^2-5(x)+(10)]}{h}$
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<div><img class="left"/><div class="alert gray">

Ok, Makes sense.

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

I am very glad it does, Would you mind working on simplifying the equation?

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

Yup! So I think simplifying would look like this:

$\lim_{h \to 0}\frac{[3x^+3h^2+6xh-5x-5h+10]-[3x^2-5x+10]}{h}$

$\lim_{h \to 0}\frac{3h^2+6xh-5h}{h}$

$\lim_{h \to 0}{3h+6x-5}$

$\lim_{h \to 0}{3(0)+6x-5}$

$= 6x-5$.
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<div><div class="alert blue">

Excellent, Yes that is exactly how you would simplify. Now Like I have said, it seemed counter-intuitive, but you actually solved for part B. This simplified equation is actually the formula for the derivative of any point in the function.

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<div><div class="alert blue">
Now having the formula to solve for the derivative(or slope) of any point, 
the derivative for when x+5 will be needed to solve for the tangent line equation(aka the Slope-intercept form).
 
$f\left(x\right)=3x^{2}-5x+10$

$x=5$

$f\left(x\right)=3(5)^{2}-5(5)+10$
$=75-25+10$

$=60$
</div><img class="right"/></div>

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

Ok, Now why do we need the slope of x=5 for the slope-intercept form?

</div></div> 

<div><div class="alert blue">
So now that we have solved for the y of the function, we now have a point in our function which we solved algarabaically: (5,60).

This point which we found can be plugged into the Slope-intercept formula as:

$60=35\left(x\right)+b$

Now when simplified, $-115=b$

In which the Tangent line formula for $x=5$ will be:
$y=35\left(x\right)-115$
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<div><img class="left"/><div class="alert gray">

A.) $y=35\left(x\right)-115$

B.) $= 6x-5$.

I see now, Makes total sense! If you don't mind I will be reviewing this for tomorrow's exam. Thanks so much man! You are much help.
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<div><div class="alert blue">

No, pronlem. Thanks again for reaching out. Good Luck in your Exam tomorrow!

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