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Little confused on two concepts. a) how a specific problem gets simplified and b)how to actually solve.

here is the problem:

$$y=-4-8x^2; P(-2,-36)$$

we get:

$$Secant Slope = \Delta y/ \Delta x = ((-4-8)-2+h^2)-(-4-8(-2)^2))/h$$

and in my textbook they simplify it to this, but don't show why! I am confused (question(a)):

$$Secant slope = \Delta y/ \Delta x = 32h - 8h^2/h$$

I can see where the $-8h^2$ comes from, but can't seem to figure out how it gets to $32h$

question (b)

My calculus teacher really doesn't explain things well, she went over how to solve this but I still don't understand. Can someone explain how to solve? This isn't a homework question by the way this is truly for my own understanding.

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It looks more from what's given that you are trying to calculate the tangent line (presumably so that you can calculate the limit to get the derivative). For a secant line, you'd need two points.

Anyway, for the tangent line:

$$\frac{\Delta y}{\Delta x} = \frac{y(x+\Delta x) - y(x)}{\Delta x}$$

Then this becomes:

$$\frac{\Delta y}{\Delta x} = \left[\frac{-4-8(x+\Delta x)^2 - (-4 - 8x^2)}{\Delta x}\right] = \left[\frac{-16 x \Delta x - 8 (\Delta x)^2}{\Delta x}\right] = -16x - 8\Delta x.$$

Plugging in $x=-2$ gives

$$\frac{\Delta y}{\Delta x} = -16(-2) - 8\Delta x = 32 - 8 \Delta x.$$

Then, if we take the limit as $\Delta x$ approaches zero, the derivative is $32$.

Hope this helps!

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  • $\begingroup$ Ohh okay. Starting to make sense. One thing though that I'm confused on is how you end up with -16x-8x, where does the -16 come from? $\endgroup$ – Omeed Feb 18 '15 at 18:38
  • $\begingroup$ $-8(x+\Delta x)^2 = -8x^2 - 16x\Delta x - 8(\Delta x)^2$ $\endgroup$ – John Feb 19 '15 at 19:17

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