Just a little question i need help with.

1. The curve C is given by the equation y=mx^{3}-x^{2}+8x+2, for a constant m

A) find dy/dx

The point P lies on C, and has the x-value 5

The normal to C at P is parallel to the line given by the equation y+4x-3=0.

B)find the gradient of the curve at C at P.

Hence find:

C) (I) the value of m,

(ii) the y-value at p.

For A) i got:

dy/dx=3mx^{2}-x+8

I'm not really sure how to carry on with it now any help would be brilliant :)

Thanks.

Re: Just a little question i need help with.

a) Check your differentiation. Your second term is wrong.

b) Evaluate $\displaystyle \frac{dy}{dx}\left|_{x=5}$

c) The statement "The normal to C at P is parallel to the line given by the equation y+4x-3=0" tells us that the derivative of y with respect to x at x = 5 is the negative reciprocal of the slope of the given line. So, find the slope of the line $\displaystyle y+4x-3=0$, invert it, negate it and equate this to $\displaystyle \frac{dy}{dx}\left|_{x=5}$.

i) Solve the resulting equation for $\displaystyle m$.

ii) We are told the *x*coordinate of point *P* is 5, so to find the *y*-coordinate evaluate $\displaystyle y(5)$.

Re: Just a little question i need help with.

Thanks i was really stuck on that question :)