Question

What is the range of the function $f\left(x\right)=-\mathbf{4}{x}^2-\frac{1}{2}\mathbf{?}$

$$\begin{gathered} \mathrm{A}\left\{\mathbf{all}\mathbf{}\mathbf{integers}\mathbf{}\mathbf{less}\mathbf{}\mathbf{than}\mathbf{}\mathbf{or}\mathbf{}\mathbf{equal}\mathbf{}\mathbf{to}\mathbf{}\frac{1}{2}\right\} \\ \mathbf{B}\left\{\mathrm{all}\mathrm{nonnegative}\mathrm{integers}\right\} \\ \mathbf{C}\left\{\mathrm{all}\mathrm{real}\mathrm{numbers}\right\} \\ \mathbf{D}\left\{\mathrm{all}\mathrm{real}\mathrm{numbers}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}-\frac{}{}\right\} \end{gathered}$$

Short answer

The range of the given function is all real numbers less than or equal to $\mathbf{-}\frac{1}{2}$.

Step-by-step solution

Step 1. Identify.

The function$f\left(x\right)=-4x^2-\frac{1}{2}$ is a parabola with$a=-4,b=0$ and $c=-\frac{1}{2}$. As a is negative, the graph opens downward, and the function has a maximum value. The range is all real numbers less than or equal to the maximum value.

Step 2. Find x coordinate of vertex.

X coordinate of vertex is given by $x=\frac{-b}{2a}$. Substitute $-4$ for a, 0 for b and $-\frac{1}{2}$ for c

$$\begin{gathered} x=\frac{-0}{2\left(-4\right)} \\ =0 \end{gathered}$$

Step 3. Find the value of y.

Substitute x-coordinate of vertex into the original equation,

$$\begin{gathered} f\left(x\right)=-4x^2-\frac{1}{2} \\ =-4{\left(0\right)}^2-\frac{1}{2} \\ =-\frac{1}{2} \end{gathered}$$

Therefore, the vertex is at $\left(0,-\frac{1}{2}\right)$.

The maximum value is $-\frac{1}{2}$. The range is all real numbers less than or equal to the maximum value, or all real numbers less than or equal to $-\frac{1}{2}$. Option D is the required answer.