# Confused on secant slopes

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.

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!

• 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? – Omeed Feb 18 '15 at 18:38
• $-8(x+\Delta x)^2 = -8x^2 - 16x\Delta x - 8(\Delta x)^2$ – John Feb 19 '15 at 19:17