question:the curve y=x^3-px^2-qx+r has the turning point at(-1,10) and (3,-22).Find the values of p,q and r.
please help me..i dont know how to do.
HI
i guess you know that you need to differentiate a given function and have its equals 0 to find its turning points .
$\displaystyle \frac{dy}{dx}=3x^2-2px-q$
$\displaystyle 3x^2-2px-q=0$
then substitute -1 and 3 into the equation and you will get 2 equations . Then solve for p and q .'
For r , substitute p , q and one of the points this curve passes (in red) into the original equation (blue) .