Find the range of the quadratic function.
-$$f ( x ) = 2 x ^ { 2 } + 2 x - 8$$
A) $\left( - \infty , - \frac { 1 } { 2 } \right]$
B) $\left( - \infty , - \frac { 17 } { 2 } \right]$
C) $\left[ - \frac { 17 } { 2 } , \infty \right)$
D) $\left[ - \frac { 1 } { 2 } , \infty \right)$

Question

Quizplus · Accepted Answer

The quadratic function $f(x) = 2x^2 + 2x - 8$ opens upwards (since the coefficient of $x^2$ is positive). To find the range, we need to find the vertex of the parabola. The x-coordinate of the vertex is given by $-\frac{b}{2a} = -\frac{2}{4} = -\frac{1}{2}$. Substituting this back into the function gives the minimum value $f(-\frac{1}{2}) = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 8 = 2(\frac{1}{4}) - 1 - 8 = -\frac{17}{2}$. Since the parabola opens upwards, the range is $\left[ - \frac { 17 } { 2 } , \infty \right)$.

Find the Range of the Quadratic Function f(x)=2x2+2x−8f ( x ) = 2 x ^ { 2 } + 2 x - 8f(x)=2x2+2x−8

Find the Range of the Quadratic Function $f ( x ) = 2 x ^ { 2 } + 2 x - 8$