Now that i've got your attention, I am struggling with Differentiation.

Using your expression for dy/dx, find the co-ordinates of another point on the curve at which the tangent is parallel the the one at (2, 40).

dy/dx = 6x^2 - 30x + 42.

Gradient m at (2,40) = 6.

PLEASE HELP ME!!! I have a demon maths teacher and my homework is due tomorrow!!!

Many thanks in advance :) **

Posted Thu 24th January, 2013 @ 20:50 by

Re-Re
Parallel means that it has the same gradient.

6x^2 - 30x + 42 = 6

Then rearrange this to find the two values of x

6x^2 - 30x + 36 = 0

divide by 6 to make it easier

x^2 - 5x + 6 = 0

(x - 2)(x - 3) = 0

so x = 2 or 3

since (2,40) has already used 2, the x coordinate for the new line is 3

integrate the curve to find the equation of the curve

y = 2x^3 - 15x^2 + 42x

substitute x = 3 into the equation

y = 2(3)^3 - 15(3)^2 + 42(3)

y = 52 - 135 + 126

y = 42

coordinates = (3,42)

Hope this helps and sorry I rushed it

Answered Thu 24th January, 2013 @ 21:27 by

Joanne