this is the question:the normal to the curve y=X^3-2X^2 at the point (1,-1)passes through the point(p,2p).calculate the value of p.
someone please help me..
HI
$\displaystyle y=x^3-2x^2$
$\displaystyle \frac{dy}{dx}=3x^2-4x$ and this is the gradient of tangent
$\displaystyle =3(1)-4$
$\displaystyle =-1$
Then using this formula , $\displaystyle y-y_1=m(x-x_1)$
you will find that the equation of normal is y=x-2
so this equation passes through (p,2p)
$\displaystyle 2p=p-2$
$\displaystyle
p=-2
$