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title: My lesson Topic

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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<style>
body {
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h1 {
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    font-weight: 500;
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.right{
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<div><img class="left"/><div class="alert gray">

So... what exactly is this assignment? it looks tricky and discouraging.

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 Consider the function $f(x) = 4x^2-20x+25$
   (a) Find the formula for the derivative of this function at any point. Use only the deﬁnition of the derivative 
   (b) Find the equation for the tangent line to the graph of this function at x = 7. Use only the deﬁnition of the derivative

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

So, where should we begin?

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We'll start by writing everything we know, we know $x=7$ and that $f(x)=4x^2-20x+25$ if we plug $7$ into our equation for all values $x$. For example, $f(7)=4(7)^2-20(7)+25$ After we plug in into our calculator we find that $f(7)=71$ Which says the y value when $x=7$ is $71$ so that gives us a coordinate on the function f(x) which exists at (7,71).
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<div><img class="left"/><div class="alert gray">

Ooohhhh ok. I think i understand. So, f(x)equals the y-value for the value x. And if we plug in a value for x will give us the cooresponding y value. But is that the answer?

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

No, thats just the start. Now we know our given values for x and y, but $a$) is asking for the derivative equation for our graph. 
The derivative is equal to the slope of function. So, we are looking for $f'(x)=lim(h->0)(f(x+h)-f(x)/h)$ Where we insert f(x+h) and f(x)into our function $$f(x)=4x^2-20x+25$$ to get our f'(x) equation. 
so $$f'(x)= limh->0 [4(x^2+2xh+h^2)-20(x+h)+25]-[4x^2-20x+25]/h$$ =linh->0$$(4x^2+8hx+4h^2-20x-20h-4x^2+20x=25)/h$$ =limh->0$$(8hx+4h^2-20h)/h$$ 
=limh->0$$8x+4h-20=8x+4(0)-20$$
=$8x-20$ which will be our derivative equation for our function $$f(x)=4x^2-20x+25$$

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

Wow! I think i understand that, but that cannot be part $a$ alone? What about part $b$?

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

Yes That was only part $a$ but part $b$ is really simple beacuse we know all the information to find our tangent line. Since we know a tangent lines' equation is $$y-y1=m(x-x1)$$ we just need to input all of our information. So, x1=7 and y1=71 and if we plug x=7 into our derivative equation we will find the slope. Because the derivative is equal to a functions' slope. So, $$f'(7)=8(7)-20=36$$ and that will be our slope 'm'. Now we just plug everything in and solve for y, so $$y-71=36(x-7)$$ $$y-71=36x-252$$ $$y=36x-181$$ And that would be our tangent line.

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

I see! So we pretty just plug all the informatio we know and found into the formula and solve for y. It was pretty discourouging at first, but I think I get the gist of it. Thank you!

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