Question

Standard to vertex form -x^2-3x+8

Answer 1

The vertex form:

$f(x)=a(x-h)^2+k$

(h, k) - vertex

$f(x)=ax^2+bx+c\\\\h=\dfrac{-b}{2a},\ k=f(h)$

We have

$f(x)=-x^2-3x+8\\\\a=-1,\ b=-3,\ c=8\\\\h=\dfrac{-(-3)}{2(-1)}=\dfrac{3}{-2}=-\dfrac{3}{2}\\\\k=f\left(-\dfrac{3}{2}\right)=-\left(-\dfrac{3}{2}\right)^2-3\left(-\dfrac{3}{2}\right)+8=-\dfrac{9}{4}+\dfrac{9}{2}+8\\\\=-\dfrac{9}{4}+\dfrac{18}{4}+\dfrac{32}{4}=\dfrac{-9+18+32}{4}=\dfrac{41}{4}\\\\Answer:f(x)=-\left(x-\left(-\dfrac{2}{3}\right)\right)^2+\dfrac{41}{4}=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{41}{2}$

Answer 2

F(x) = - (x-(2/3))^2 41/2 = - (x 3/2) ^2