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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, is this Emily from Math 181? 

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

Yeah, who's this?

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


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

This is Casey, I'm the one who asked for your number after class.

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

Could you help me with problem 3?

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

Yeah of course, the one about findin the limit and a tangent line?

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

Yeah, I get that the equation line will be all over h, but I don't know what goes on too in the numerator

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

Okay, so it's great that you know in the limit definiton f(x)= lim as h approaches 0 that the divisor is h. The numerator would be f(x+h)-(f(x). This seems tricky but its not. I know you know f(x) becaue its given in the problem that f(x)=3x^2-5x+10. To getf(x+h), we fill in an x+h instead of just x so its [3(x+h)^2-5(x+h)+10] and then of course minus f(x) which we said was 3x^2-5x+10. Dont forget to distrubute the negative sign.After you simplify you should get 6x-5. Its implifies to 6x+3h+5, but since h is 0 we dont need it.
(http://sergeballif.github.io/NSC-Math-181/Specifications.html).


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

I ended up simplifying and I got 6h-5. We use this to find the tangent line right? When x=5? Do we just fill in 5 for x. I forgot the formula.

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

So we actually use a diferent forumla first but we plug in five there. ITs the formula y=f(5)+f'(5)(x-5). Ill show you my work
=[3(5^2-5(5)+10)]+[6(5)-5](x-5)
=75-25+10(x-5)
=60+25(x-5)
Hope this helped!!

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


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