Chapter 3, Problem 22RE

(a)

To determine

To graph: the given quadratic function

Answer to Problem 22RE

  

Explanation of Solution

Given:

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

Calculation:

(a) graph the function $f\left(x\right)=-x^2+x+\frac{1}{2}$ . 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.

  $\begin{array}{c}-x^2+x+\frac{1}{2}=0\\ x=\frac{-1\pm \sqrt{1+2}}{-2}\\ x=\frac{-1\pm \sqrt{3}}{-2}\\ x=\frac{-1+\sqrt{3}}{-2}\text{or}\\ x=\frac{-1-\sqrt{3}}{-2}\\ x=\frac{1-\sqrt{3}}{2}\text{or}\\ x=\frac{1+\sqrt{3}}{2}\end{array}$

The vertex is at

  $\begin{array}{c}x=-\frac{1}{2\left(-1\right)}\\ =\frac{1}{2}\end{array}$

Since $f\left(\frac{1}{2}\right)=\frac{3}{4}$ , thus the vertex is $\left(\frac{1}{2},\frac{3}{4}\right)$

  

The graph of $f\left(x\right)=-x^2+x+\frac{1}{2}$ is a parabola that opens down and has it vertex at $\left(\frac{1}{2},\frac{3}{4}\right)$ .

Its axis of symmetry is the line $x=\frac{1}{2}$ . Its y-intercept is 2 and x-intercepts are $x=\frac{1-\sqrt{3}}{2}$ and $x=\frac{1+\sqrt{3}}{2}$

Conclusion:

Thus, the given equation is drawn.

(b)

To determine

the domain and the range of the function.

Answer to Problem 22RE

The domain is $\left(-\infty ,\infty \right)$ .

The range is $\left(-\infty ,\frac{3}{4}\right)$ .

Explanation of Solution

Calculation:

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

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 $f\left(x\right)=-x^2+x+\frac{1}{2}$ shows that the vertex $\left(\frac{1}{2},\frac{3}{4}\right)$ of the function is a maximum point.

So, its range is $\left(-\infty ,\frac{3}{4}\right)$ .

Conclusion:

The domain is $\left(-\infty ,\infty \right)$ .

The range is $\left(-\infty ,\frac{3}{4}\right)$ .

(c)

To determine

the increasing and decreasing interval.

Answer to Problem 22RE

Decreasing interval $\left(\frac{1}{2},\infty \right)$ and increasing interval $\left(-\infty ,\frac{1}{2}\right)$

Explanation of Solution

Calculation:

Determine whether the function $f\left(x\right)=-x^2+x+\frac{1}{2}$ is increasing, decreasing orconstant. The graph of the function $f\left(x\right)=-x^2+x+\frac{1}{2}$ shows that the vertex $\left(\frac{1}{2},\frac{3}{4}\right)$ of the function is a maximum point. Hence, decreasing interval $\left(\frac{1}{2},\infty \right)$ and increasing interval $\left(-\infty ,\frac{1}{2}\right)$

Conclusion:

Hence, decreasing interval $\left(\frac{1}{2},\infty \right)$ and increasing interval $\left(-\infty ,\frac{1}{2}\right)$