Question

Find the equation of the tangent line to F(x)=(x-8)^4 at the point where x = 2

Answer 1

So you need to use chain rule to find F'

Answer 2

Have you done that?

Answer 3

yea i got 4(x-8)^3

Answer 4

ok great now find F'(2) to find the slope at (2,F(2))

Answer 5

not sure how to set that equation up

Answer 6

You told me F'(x)
which is 4(x-8)^3
So we have
F'(x)=4(x-8)^3
Since you want to find F'(2)
you just replace x with 2 in F'(x)=4(x-8)^3
like so
F'(2)=4(2-8)^3=....

Answer 7

\[F'(2)=4(2-8)^3=4(-6)^3=-4(6)^3\]
-4(6)^3 is the slope at (2,F(2))
And by the way F(2)=(2-8)^4=(-6)^4=6^4

Answer 8

\[y=mx+b \]
is the form in which all lines can be written
for this line we have m=-4(6)^3 and the point in which the line goes through which is
(2,6^4)

Answer 9

\[6^4=-4(6)^3 \cdot 2+b\]
I replaced (x,y) with (2,6^4) and m with -4(6)^3 to find the y-intercept of the tangent line b

Answer 10

\[y=-4(6)^3x+(6^4+4(6)^3 \cdot 2)\]