Chapter 3, Problem 15RE

(a)

To determine

To graph: the given quadratic function

Answer to Problem 15RE

  

Explanation of Solution

Given:

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

Calculation:

Graph the function $f\left(x\right)={\left(x-2\right)}^2+2$ . To find the y-intercept, evaluate the functionat $x=0$ , y-intercept: $f\left(0\right)=2$

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

  $\begin{array}{c}{\left(x-2\right)}^2+2=0\\ x^2-4x+4+2=0\\ x^2-4x+6=0\\ x=\frac{4\pm \sqrt{16-24}}{2}\\ x=\frac{4\pm \sqrt{-8}}{2}\\ x_1=\frac{2+2\sqrt{2}i}{2}\\ =1+i\sqrt{2}\\ x_2=\frac{2-2\sqrt{2}i}{2}\\ =1-i\sqrt{2}\end{array}$

No real x-intercept is found. So, the function $f\left(x\right)={\left(x-2\right)}^2+2$ has imaginary intercept or no real intercept. Its discriminant $b^2-4ac=16-4\left(1\right)\left(6\right)=-8<0$ , so it does not cross or touch the x-axis.

The vertex is at

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{-4}{2}\\ =2\end{array}$

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

  

The graph of $y=f\left(x\right)={\left(x-2\right)}^2+2$ is a parabola that opens up and has its vertex (lowest point) at $\left(2,2\right)$ . Its axis of symmetry is the line $x=2$ . Its y-intercept is 6 and it has no real x-intercept.

Conclusion:

Thus, the given equation is drawn.

(b)

To determine

the domain and the range of the function.

Answer to Problem 15RE

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

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

Explanation of Solution

Calculation:

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

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 minimum value of function is $\left(2,2\right)$ . So, its range is $\left(2,\infty \right)$ .

Conclusion:

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

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

(c)

To determine

the increasing and decreasing interval.

Answer to Problem 15RE

Hence, decreasing interval $\left(-\infty ,2\right)$ and increasing interval $\left(2,\infty \right)$ .

Explanation of Solution

Calculation:

Determine whether the function $f\left(x\right)={\left(x-2\right)}^2+2$ is increasing, decreasing or constant. The graph of the function $f\left(x\right)={\left(x-2\right)}^2+2$ shows that the vertex $\left(2,2\right)$ of the function is a minimum point. So, the function $f\left(x\right)={\left(x-2\right)}^2+2$ is decreasing from $-\infty$ to vertex and increasing from vertex to $\infty$ .

Conclusion:

Hence, decreasing interval $\left(-\infty ,2\right)$ and increasing interval $\left(2,\infty \right)$ .