Find the vertex
-$f(x)=x^{2}-8 x+24$
A) $(0,4)$
B) $(4,8)$
C) $(8,0)$
D) $(8,4)$

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 $f(x) = x^2 - 8x + 24$, $a = 1$ and $b = -8$, so the x-coordinate of the vertex is $-(-8)/(2*1) = 4$. Plugging $x = 4$ back into the equation gives $f(4) = (4)^2 - 8(4) + 24 = 16 - 32 + 24 = 8$. Therefore, the vertex is $(4, 8)$.

Find the Vertex - f(x)=x2−8x+24f(x)=x^{2}-8 x+24f(x)=x2−8x+24 A) (0,4)(0,4)(0,4) B) (4,8)(4,8)(4,8) C) (8,0)(8,0)(8,0) D) (8,4)(8,4)(8,4)

Find the Vertex
- $f(x)=x^{2}-8 x+24$ A) $(0,4)$ B) $(4,8)$ C) $(8,0)$ D) $(8,4)$