Find the coordinates of the vertex and the intercepts of the quadratic function. When necessary, approximate the
x-intercepts to the nearest tenth.
-$$f ( x ) = - 4 x ^ { 2 } - 2 x - 6$$
A) V(4,-6); I(0, -6); no $x$-intercepts
B) $\mathrm { V } \left( - 4 , - \frac { 23 } { 4 } \right) ; \mathrm { I } ( 0,6 ) ; ( - 0.9,0 ) , ( 1.4,0 )$
C) $\mathrm { V } \left( \frac { 1 } { 4 } , \frac { 23 } { 4 } \right) ; \mathrm { I } ( 0,6 ) ; ( - 1.4,0 ) , ( 0.9,0 )$
D) $\mathrm { V } \left( - \frac { 1 } { 4 } , - \frac { 23 } { 4 } \right) ; \mathrm { I } ( 0 , - 6 )$; no $x$-intercepts

Question

Quizplus · Accepted Answer

The vertex of a quadratic function $f(x) = ax^2 + bx + c$ can be found using the formula $(-b/2a, f(-b/2a))$. For the given function, $a = -4$, $b = -2$, and $c = -6$, so the vertex is at $(-(-2)/(2*(-4)), f(-(-2)/(2*(-4)))) = (-1/4, -23/4)$. The $y$-intercept is found by evaluating $f(0)$, which gives $-6$. To find $x$-intercepts, we solve $f(x) = 0$, but the discriminant $b^2 - 4ac = (-2)^2 - 4(-4)(-6) = 4 - 96 = -92$ is negative, indicating there are no real $x$-intercepts.

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