Chapter 3, Problem 16RE

(a)

To determine

To graph: the given quadratic function

Answer to Problem 16RE

  

Explanation of Solution

Given: $f\left(x\right)={\left(x+1\right)}^2-4$

Calculation:

Graph the function $f\left(x\right)={\left(x+1\right)}^2-4$ . To find the y-intercept, evaluate the functionat $x=0$ , y-intercept: $f\left(0\right)=-3$

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

  $\begin{array}{c}{\left(x+1\right)}^2-4=0\\ x^2+2x+2-4=0\\ x^2-2x-2=0\\ x^2+2x-x-2=0\\ x\left(x+2\right)-1\left(x+2\right)=0\\ \left(x+2\right)\left(x-1\right)=0\\ x+2=0\text{or}\\ x-1=0\\ x=2\text{or}\\ x=1\end{array}$

The vertex is at

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{2}{2}\\ =-1\end{array}$

Since $f\left(-1\right)=-4$ , thus the vertex is $\left(-1,-4\right)$ .

  

The graph of $f\left(x\right)={\left(x+1\right)}^2-4$ is a parabola that opens up and has its vertex (lowest point) at $\left(-1,-4\right)$ . Its axis of symmetry is the line $x=-1$ . Its y-intercept is $x=-3$ and x-intercepts are 1 and 2.

Conclusion:

Thus, the given equation is drawn.

(b)

To determine

the domain and the range of the function.

Answer to Problem 16RE

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

The range is $\left(-1,\infty \right)$

Explanation of Solution

Calculation:

Determine the range and domain of the function $f\left(x\right)={\left(x+1\right)}^2-4$ .

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

So, its range is $\left(-1,\infty \right)$ .

Conclusion:

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

The range is $\left(-1,\infty \right)$

(c)

To determine

the increasing and decreasing interval.

Answer to Problem 16RE

Hence, decreasing interval and increasing interval $\left(-1,\infty \right)$ .

Explanation of Solution

Calculation:

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

Conclusion:

Hence, decreasing interval and increasing interval $\left(-1,\infty \right)$ .