x Find the range of the quadratic function.
-$$f ( x ) = 4 x ^ { 2 } + 2 x - 6$$
A) $\left[ - \frac { 25 } { 4 } , \infty \right)$
B) $\left( - \infty , - \frac { 25 } { 4 } \right]$
C) $\left[ - \frac { 1 } { 4 } , \infty \right)$
D) $\left( - \infty , - \frac { 1 } { 4 } \right]$

Question

Quizplus · Accepted Answer

The quadratic function $f(x) = 4x^2 + 2x - 6$ opens upwards (since the coefficient of $x^2$ is positive). To find the range, we need to find the vertex of the parabola, which gives the minimum value of the function. The vertex formula is $(-b/2a, f(-b/2a))$. Here, $a = 4$ and $b = 2$, so the x-coordinate of the vertex is $-2/(2*4) = -1/4$. Plugging this back into the function gives the minimum value: $f(-1/4) = 4(-1/4)^2 + 2(-1/4) - 6 = 1 - 1/2 - 6 = -25/4$. Therefore, the range is $\left[ - \frac { 25 } { 4 } , \infty \right)$.

X Find the Range of the Quadratic Function f(x)=4x2+2x−6f ( x ) = 4 x ^ { 2 } + 2 x - 6f(x)=4x2+2x−6

X Find the Range of the Quadratic Function $f ( x ) = 4 x ^ { 2 } + 2 x - 6$