Chapter 3, Problem 18RE

(a)

To determine

To graph: the given quadratic function

Answer to Problem 18RE

  

Explanation of Solution

Given:

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

Calculation:

graph the function $f\left(x\right)=-\frac{1}{2}{x}^2+2$ . To find the y-intercept, evaluate the function

at $x=0$ , y-intercept: $f\left(0\right)=2$

Now, solve $f\left(x\right)=0$ , to find the x-intercepts.

  $\begin{array}{c}-\frac{1}{2}{x}^2+2=0\\ -\frac{1}{2}{x}^2=-2\\ x^2=4\\ x=\pm \sqrt{4}\\ x=\pm 2\\ x=2\text{or}\\ x=-2\end{array}$

The vertex is at

  $\begin{array}{c}x=-\frac{b}{2a}\\ =0\end{array}$

Since $f\left(0\right)=2$ , thus the vertex is $\left(0,2\right)$ .

  

The graph of $f\left(x\right)=-\frac{1}{2}{x}^2+2$ is a parabola that opens down and has its vertex at $\left(0,2\right)$ . Its axis of symmetry is the line $x=0$ . Its y-intercept is 2 and x-intercepts are $-2$ and 2.

Conclusion:

Thus, the given equation is drawn.

(b)

To determine

the domain and the range of the function.

Answer to Problem 18RE

The domain of quadratic equation is all real numbers $\left(-\infty ,\infty \right)$ .

The range is $\left(-\infty ,2\right)$

Explanation of Solution

Calculation:

Determine the range and domain of the function $f\left(x\right)=-\frac{1}{2}{x}^2+2$ .

The domain of quadratic equation is all real numbers $\left(-\infty ,\infty \right)$ .

For the range, find the minimum and maximum value of function. The graph of the function $f\left(x\right)=-\frac{1}{2}{x}^2+2$ shows that the vertex $\left(0,2\right)$ of the function is a maximum point. So,its range is $\left(-\infty ,2\right)$ .

Conclusion:

The domain of quadratic equation is all real numbers $\left(-\infty ,\infty \right)$ .

The range is $\left(-\infty ,2\right)$

(c)

To determine

the increasing and decreasing interval.

Answer to Problem 18RE

Decreasing interval $\left(0,\infty \right)$ and increasing interval $\left(-\infty ,0\right)$

Explanation of Solution

Calculation:

Determine whether the function $f\left(x\right)=-\frac{1}{2}{x}^2+2$ is increasing, decreasing orconstant. The graph of the function $f\left(x\right)=-\frac{1}{2}{x}^2+2$ shows that the vertex $\left(0,2\right)$ of the function is a maximum point. Hence, decreasing interval $\left(0,\infty \right)$ and increasing interval $\left(-\infty ,0\right)$

Conclusion:

Hence, decreasing interval $\left(0,\infty \right)$ and increasing interval $\left(-\infty ,0\right)$