# Slope of a tangent

1. Jul 5, 2006

### richievuong

Sry dont know how to use latex, so my calculations may be very messy

Find the coordinates of the point on the curve f(x) = 3x^2-4x , where the tangent is parallel to the line y=8x.

I know the formula m=limh->0 f(x+h)-f(x) / h

What i tried was:

Let coordinates be (x,y)

y=3x^2-4x
f(x) = y
f(x+h) = 3(x+h)^2 - 4(x+h)
= 3x^2+6xh+3h^2-4x-4h

m=limh->0 f(x+h)-f(x) / h
m=limh->0 3x^2+6xh+3h^2-4x-4h-(3x^2-4x) / h
m=limh->0 3^2+6xh+3h^2-4x-4h-3x^2+4x / h
m=limh->0 6xh+3h^2-4h / h

Since tangent line has to be parallel to y=8x
8 = 6xh+3h^2-4h / h
8h = 6xh+3h^2-4h
12h = 6xh+3h^2

Divide by h on both sides:

12 = 6x + 3h

Sub 0 for h

12=6x
x=2

y=3x^2-4x
y=3(2)^2-4(2)
y=12-8
y=4

Therefore the coordinates on the curve that make the tangent line parallel to y=8x is (2,4)

Please check over my work, I think I did something wrong.

2. Jul 5, 2006

### neutrino

Your answer is correct, although it would have been prettier if you had evaluated the limit first.