Question

Find the equation for the normal line to f at the point (-2, f(-2)).
f(x)= -x +4x^2

Answer 1

Have you considered the 1st Derivative as a clue to find the slope of the Normal Line?

Answer 2

the first derivate is 8x

Answer 3

Evaluated at x = -2. What is the slope of the tangent line at that point?

Answer 4

-16?

Answer 5

Good. Now, what is the slope of the NORMAL line at that point?

Answer 6

18!

Answer 7

sorry 14?

Answer 8

so is y -18 = 16x?

Answer 9

?? The Normal is Perpendicular to teh Tangent. Give it another go.

Answer 10

wait!, okay the derivative for f(x) = -x +4x^2 is equal to 8x-1, thus the answer is -17.
okay after it whats next?

Answer 11

0.o?

Answer 12

Your question is not making sense? Or am i reading it wrong.

Answer 13

Whoops! Missed the x. Good catch.

f(x) = -x + 4x^2
f(-2) = 2 + 4(4) = 2+16 = 18
f'(x) = -1 + 8x
f'(-2) = -1 - 16 = -17

The tangent line at (-2,18) is (y-18) = -17(x+2)
The normal line is perpendicular to the tangent line and passes through the same point. You tell me what that equation is.

Answer 14

answer*** not question.

Answer 15

where the (x+2) come from?

Answer 16

You should know the Point-Slope form of a line.

Point (a,b)
Slope m

(y-b)=m(x-a)

Don't forget your algebra!!

Answer 17

M.A.T.H stand for Mental Abuse to Humans, my brain does not have enough memory for more!

Answer 18

Blah, blah. You can do it. Hang in there! :-)