Question

Find the equation of a normal line to the given curve with the given slope

y=(4x-7)^3, slope -1/12, x>0

So far, I have:

y'=12(4x-7)^2
12(4x-7)^2=12
4x-7=+1,-1
x=2, x=3/2

Answer 1

well the slope of our line must be -1/(-1/12) = 12 agree

I have y' = 3 (4x-7)^2 (4) = 12 (4x-7)^2 agree
that is at (4x-7)^2 = 1
4x - 7 = +1
x = 2
or
x = 3/2 agree
let's use x = 2
then y = (4*2-7)^3 = 1
so through the point (2 , 1)
y = (-1/12) x + b
1 = (-1/12) (2) + b
b = 1 + 1/6 = 7/6 = 14/12
y = - x/12 + 14/12
12 y = -x + 14

so
y = 12 x -23

Answer 2

forget that last line, error