Question

A. Find the derivative of f at x. That is, find f'(x).
B. Find the slope of the tangent line to the graph of f at each of the two values of x given to the right of the function

f(x)=x^2-8; x=-1, x=3

Answer 1

@ganeshie8

Answer 2

HI!!

Answer 3

Hey

Answer 4

you know the derivative of \(x^2\)?

Answer 5

if the answer is "no" that is fine, i will tell you

Answer 6

is it 2x?

Answer 7

yes

Answer 8

and the derivative of a constant is zero, making the derivative of \(f(x)=x^2-8\) the function
\[f'(x)=2x\] that is all

Answer 9

but its two parts to the question is 2x the answer to them both

Answer 10

@misty1212

Answer 11

no \(2x\) is the derivative
you now need the equation for two lines, one where \(x=-1\) and the other where \(x=-3\)

Answer 12

f(x)=-1^2-8 and f(x)=-3^2-8

Answer 13

if \(x=-1\) then \(y=(-1)^2-8=-7\) so the point is \((-1,-7)\) and the slope is \(2\times -1=-2\) use the point slope formula to get the equation of the line

Answer 14

if \(x=3\) then \(y=3^2-8=1\) the point is \((3,1)\) and the slope is \(2\times 3=6\) points slope again

Answer 15

y-(-7)=-2(x-(-1)) and
y-(1)=6(x-(3))

Answer 16

@misty1212

Answer 17

you can clean them up a great deal, but yes

Answer 18

y=-2x+5
y=6x-17

Answer 19

@misty1212

Answer 20

o Thank you