Chapter 3, Problem 17RE

(a).

To determine

To graph: Quadratic function with its vertex, axis of symmetry, $y-$ intercept or $x-$ intercept.

Answer to Problem 17RE

Function using the vertex and the $x-$ intercept graph is:

  

Explanation of Solution

Given:

  $f\left(x\right)=\frac{1}{4}{x}^2-16$

Calculation:

The quadratic function $\frac{1}{4}{x}^2-16$ is form $f\left(x\right)=a{x}^2+bx+c$ .

So, $a=\frac{1}{4},b=0,c=-16$

Here, $a=\frac{1}{4}>0$

So the graph opens up.

To find the vertex,

The coordinates of the vertex of a quadratic function of the form $f\left(x\right)=a{x}^2+bx+c$ is $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$ .

Evaluating $\frac{-b}{2a}$ to find the $x-$ coordinate of the vertex. Substitute the values $a=\frac{1}{4},b=0$ in $\frac{-b}{2a}=0$

Evaluating $f$ at $\left(\frac{-b}{2a}\right)$ to find y-coordinate of the vertex. So put $0$ for $\left(\frac{-b}{2a}\right)$ and evaluate the function.

  $\begin{array}{l}f\left(\frac{-b}{2a}\right)=f\left(0\right)\\ =0\end{array}$

Thus, the vertex is at $\left(0,0\right)$ .

The axis of symmetry is the line where $x=\frac{-b}{2a}$

Thus, $x=0$ is the axis of symmetry.

To find $y-$ intercept,

  $y-$ Intercept is the value of $f$ at $x=0$

Here, $f\left(0\right)=0^2-16=-16$

Therefore, the graph has the $y-$ intercept at $-16$ .

Find the discriminant $b^2-4ac$ of the equation by substituting $a=\frac{1}{4},b=0,c=-16$ .

  $\begin{array}{l}{b}^2-4ac=0-4\left(\frac{1}{4}\right)\left(-16\right)\\ =16\end{array}$

In case discriminant $b^2-4ac>0$ , the graph of $f\left(x\right)=a{x}^2+bx+c$ has two distinct $x-$ intercepts.

Therefore, the graph of the function crosses the $x-$ axis in two places.

To find out the $x-$ intercept,

The $x-$ intercept can be obtained by solving the quadratic equation of the from $a{x}^2+bx+c=0$

  $\frac{1}{4}{x}^2-16$

Multiplying each side of the equation by $4$ .

  $\begin{array}{l}\frac{4}{4}{x}^2-64=0\\ x^2-64=0\end{array}$

Add $64$ to the both sides,

  $\begin{array}{l}{x}^2-64+64=0+64\\ x^2=64\end{array}$

Make use of square root method to solve the equation. According to the square root method, if

  $x^2=r$ and $r\ge 0$ , then $x=\sqrt{r}$ or $x=-\sqrt{r}$

  $\begin{array}{l}{x}^2=64\\ x=\pm \sqrt{64}\\ x=8orx=-8\end{array}$

Function using the vertex and the $x-$ intercept graph is:

  

Conclusion:

Quadratic function with its vertex is $\left(0,-16\right)$ , axis of symmetry, $y-$ intercept or $x-$ intercept.

(b).

To determine

Domain and range of the function.

Answer to Problem 17RE

Domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left[-16,\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{1}{4}{x}^2-16$

Calculation:

Find the domain and range of the function, the domain of $f$ is the set of all real numbers.

From the graph, find the domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left[-16,\infty \right)$ .

Conclusion:

Thus, domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left[-16,\infty \right)$ .

(c).

To determine

Determine the function is increasing and where it is decreasing.

Answer to Problem 17RE

The function $f$ is decreasing on the interval $\left(-\infty ,0\right)$ and increasing on the interval $\left(0,\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{1}{4}{x}^2-16$

Calculation:

Interval where the function $f$ is decreasing.

Also,, find where the function is increasing.

From the graph, the function $f$ is decreasing on the interval $\left(-\infty ,0\right)$ and increasing on the interval $\left(0,\infty \right)$ .

Conclusion:

Thus, the function $f$ is decreasing on the interval $\left(-\infty ,0\right)$ and increasing on the interval $\left(0,\infty \right)$ .