Chapter 8, Problem 7CT

To determine

To graph: The quadratic function $f\left(x\right)=16x^2-4$ and evaluate the domain and range of the function.

Explanation of Solution

Given information:

The equation of quadratic function $f\left(x\right)=16x^2-4$ .

Graph:

Consider the provided function $f\left(x\right)=16x^2-4$ . The function is of the form $f\left(x\right)=a{x}^2+bx+c$ where a is positive. Here $a=16$ , the function is a parabola that open upward since $a>0$ . Also, the c represents the y-intercept of the parabola.

Therefore, the graph touches the y-axis at the point $\left(0,-4\right)$ .

Now, factor the function,

  $\begin{array}{c}f\left(x\right)=16x^2-4\\ ={\left(4x\right)}^2-2^2\\ =\left(4x-2\right)\left(4x+2\right)\end{array}$

Now, the zeroes of the function is when $f\left(x\right)$ is equated to 0.

Therefore, $\left(4x-2\right)\left(4x+2\right)=0$ . Either $\left(4x-2\right)=0\text{ or }\left(4x+2\right)=0$ .

When $\left(4x-2\right)=0$ , simplify the expression,

  $\begin{array}{c}4x-2=0\\ 4x=2\\ x=\frac{2}{4}\\ x=\frac{1}{2}\end{array}$

When $\left(4x+2\right)=0$ , simplify the expression,

  $\begin{array}{c}4x+2=0\\ 4x=-2\\ x=-\frac{2}{4}\\ x=-\frac{1}{2}\end{array}$

Now, the zeroes of polynomial are $\frac{1}{2},-\frac{1}{2}$ . So, the parabola passes through the point $\left(\frac{1}{2},0\right)\text{ and }\left(\frac{-1}{2},0\right)$ .

Plot the ordered pairs on a coordinate plane and join the points to obtain the smooth curve.

The red curve represents the quadratic function $f\left(x\right)=16x^2-4$ .

  

Interpretation:

Domain of the function is the set of values of independent variable for which function is defined.

The range of the function is the set of values of the dependent variable.

The quadratic functions $f\left(x\right)=16x^2-4$ is a polynomial function and polynomial functions are defined on the entire number line. Therefore, domain of the function is $\left(-\infty ,\infty \right)$ . Now, the range of the function is the set of all values of y such that y is greater than or equal to $-4$ . The same can be observed from the graph. Therefore, range of the function is $\left[-4,\infty \right)$ .