Sry dont know how to use latex, so my calculations may be very messy(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Slope of a tangent

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