Find the vertex
-$f(x)=4 x^{2}-40 x+104$
A) $(5,4)$
B) $(-5,-4)$
C) $(4,5)$
D) $(-4,-5)$

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 $f(x) = 4x^2 - 40x + 104$, $a = 4$ and $b = -40$, so the x-coordinate of the vertex is $-(-40)/(2*4) = 5$. Plugging $x = 5$ back into the function gives $f(5) = 4(5)^2 - 40(5) + 104 = 4$. Therefore, the vertex is $(5, 4)$.

Find the Vertex - f(x)=4x2−40x+104f(x)=4 x^{2}-40 x+104f(x)=4x2−40x+104 A) (5,4)(5,4)(5,4) B) (−5,−4)(-5,-4)(−5,−4) C) (4,5)(4,5)(4,5) D) (−4,−5)(-4,-5)(−4,−5)

Find the Vertex
- $f(x)=4 x^{2}-40 x+104$ A) $(5,4)$ B) $(-5,-4)$ C) $(4,5)$ D) $(-4,-5)$