1. $\displaystyle y=(1-x^2)(4-x^2)$
2. $\displaystyle y=3x^4-16x^3+18x^2$
Answer:
1. $\displaystyle (-\sqrt{\frac{5}{2}}, -2) (\sqrt{\frac{5}{2}}, -2)$
2. $\displaystyle (1, 5) (3, -27)$
The only way I know to solve this question is to employ some calculus and differentiate. Differentiate to find $\displaystyle \frac{dy}{dx}$, then set $\displaystyle \frac{dy}{dx}=0$, and solve the resulting equation.
Eg. No.2
$\displaystyle \frac{dy}{dx}=12x^3-48x^2+36x$
$\displaystyle =12x(x^2-4x+3)$
$\displaystyle \frac{dy}{dx}=12x(x^2-4x+3)=0$
$\displaystyle =(12x)(x-1)(x-3)=0$
$\displaystyle x=0,1,3$
Place these results into the original equation
When $\displaystyle x=1$
$\displaystyle y=3(1)^4-16(1)^3+18(1)^2=5$
When$\displaystyle x=3$
$\displaystyle y=3(3)^4-16(3)^3+18(3)^2=-27$
When$\displaystyle x=0,y=0$
So your turning points are
$\displaystyle (0,0)$
Do the same for no.1