Find the relative extrema of f(x) = 1+8x − 3x^2

F(x)=1+8x-3x²
1)we have to calculate the first derived
f´(x)=8-6x
2) we have to equalizate the first derived to "0", and find out the value of "x".
8-6x=0
-6x=-8
x=-8/-6=4/3
3)we have to calculate the second derived
f´´(X)=-6<0  ⇒ at x=4/3 we have a maximum.
4) we find the value of "y"
f(4/3)=1+8(4/3)-3(4/3)²=1+(32/3)-(48/9)=57/9

Therefore:
at (4/3 , 57/9) have a maximum