Find the coordinates of the vertex and the intercepts of the quadratic function. When necessary, approximate the
x-intercepts to the nearest tenth.
-$$f ( x ) = x ^ { 2 } - 8 x + 12$$
A) $\mathrm { V } ( 4 , - 2 ) ; \mathrm { I } ( 0,6 ) ; ( 12,0 ) ; ( - 12,0 )$
B) $\mathrm { V } ( - 4 , - 4 ) ; \mathrm { I } ( 0 , \quad$; no $\mathrm { x }$-intercepts
C) $\mathrm { V } ( 4 , - 4 )$; I $( 0,12 ) ; ( 6,0 ) ; ( 2,0 )$
D) $\mathrm { V } ( 4,2 )$; I( $0 , - 12 ; ( - 2,0 ) ; ( - 6,0 )$

Question

Quizplus · Accepted Answer

The vertex of a quadratic function in the form $f(x) = ax^2 + bx + c$ can be found using the formula $(-b/2a, f(-b/2a))$. For $f(x) = x^2 - 8x + 12$, the vertex is at $x = 8/2 = 4$, and $f(4) = 4^2 - 8*4 + 12 = -4$, so the vertex is $V(4, -4)$. The y-intercept is found by setting $x = 0$, giving $f(0) = 12$, so $I(0, 12)$. To find the x-intercepts, solve $x^2 - 8x + 12 = 0$, which factors to $(x-6)(x-2) = 0$, giving x-intercepts at $x = 6$ and $x = 2$, or $I(6,0); (2,0)$.

Find the Coordinates of the Vertex and the Intercepts of the Quadratic