First integrate f ''(x), so that:
f'(x)=8x + C
Now, substitute that f '(1)=0, so that: Is this correct in the question?
f'(0)=8(1)+C=0
C=-8
Integrate again:
f(x)=4x^2-8x+C
When \ y=f(x)=3, x=1
y=3=4(1)^2 -8(1)+ C
C=7
So now we get the equation:
y=4x^2-8x+7