Find the vertex of the parabola
-$f(x)=-3 x^{2}+6 x-8$
A) $(-1,5)$
B) $(1,-5)$
C) $(-5,1)$
D) $(5,-1)$

Question

Quizplus · Accepted Answer

The vertex of a parabola in the form $f(x) = ax^2 + bx + c$ can be found using the formula $(-b/2a, f(-b/2a))$. For $f(x)=-3x^2+6x-8$, $a = -3$, $b = 6$, so the x-coordinate of the vertex is $-b/2a = -6/(2*-3) = 1$. Plugging $x = 1$ into $f(x)$ gives $f(1) = -3(1)^2 + 6(1) - 8 = -5$. Therefore, the vertex is $(1, -5)$.

Find the Vertex of the Parabola - f(x)=−3x2+6x−8f(x)=-3 x^{2}+6 x-8f(x)=−3x2+6x−8 A) (−1,5)(-1,5)(−1,5)

Find the Vertex of the Parabola
- $f(x)=-3 x^{2}+6 x-8$ A) $(-1,5)$