Q77. - Algebra 1

Question

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

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

Step-by-Step Solution

Answer

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

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.