Chapter 3.1, Problem 16E

a.

To determine

To Express: The quadratic function in standard form.

Answer to Problem 16E

the quadratic function is expressed in standard form as $f\left(x\right)=-1{\left(x+2\right)}^2+8$

Explanation of Solution

Given: The function is $f\left(x\right)=-x^2-4x+4$

Calculation:

The quadratic function $f\left(x\right)=-x^2-4x+4$ is expressed in standard form as:

  $f\left(x\right)=a{\left(x-h\right)}^2+k$ , by completing the square. The graph of the function f is a parabola with vertex $\left(h,k\right)$

The parabola opens downwards if $a<0$ .

Solve the function:

  $\begin{array}{l}f\left(x\right)=\left(-x^2-4x+4\right)\\ f\left(x\right)=-1\left(x^2+4x\right)+4\text{ [factor }-\text{1 from the }x\text{ terms]}\\ f\left(x\right)=-1\left(x^2+4x+4\right)+4-\left(-\text{1}\cdot 4\right)\text{ [complete the square : add 4 inside parentheses, subtract }\left(-\text{1}\cdot 4\right)\text{ outside]}\\ f\left(x\right)=-1{\left(x+2\right)}^2+8\text{ [factor and multiply]}\end{array}$

On comparing the above equation with standard form $f\left(x\right)=a{\left(x-h\right)}^2+k$ ,

Therefore, the quadratic function is expressed in standard form as $f\left(x\right)=-1{\left(x+2\right)}^2+8$

b.

To determine

To Find: The vertex, $x-\text{intercept}$ and $y-\text{intercept}$

Answer to Problem 16E

the vertex is $\left(h,k\right)=\left(-2,8\right)$ .

  $x\text{-intercept}$ is $2\pm \sqrt{6}$

  $y\text{-intercept}=f\left(0\right)=4$

Explanation of Solution

Given: The function is $f\left(x\right)=-x^2-4x+4$

Calculation:

The quadratic function $f\left(x\right)=-x^2-4x+4$ is expressed in standard form as:

  $f\left(x\right)=-1{\left(x+2\right)}^2+8$

by completing the square. The graph of the function f is a parabola with vertex $\left(h,k\right)$ , the vertex is $\left(h,k\right)=\left(-2,8\right)$ .

The $y\text{-intercept}$ is given when $x=0$ , so $y\text{-intercept}=f\left(0\right)=4$

Use Quadratic formula to find the $x\text{-intercept}$ :

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{Here; }a=-1,b=-4,c=2\\ x=\frac{4\pm \sqrt{{\left(-4\right)}^2-4\left(-1\right)2}}{2\left(-1\right)}\\ x=\frac{4\pm \sqrt{16+8}}{-2}\\ x=\frac{4\pm \sqrt{24}}{-2}=2\pm \sqrt{6}\end{array}$

The $x\text{-intercept}$ is $2\pm \sqrt{6}$

c.

To determine

To Sketch: The graph of the quadratic function.

Explanation of Solution

Given: The function is $f\left(x\right)=-x^2-4x+4$

Graph:

The standard form of the function is:

  $f\left(x\right)=-1{\left(x+2\right)}^2+8$

From the standard form it is observed that the graph is a parabola that opens downward and has vertex $\left(-2,8\right)$ . As an aid to sketching the graph, find the intercepts. The $y\text{-intercept}=f\left(0\right)=4$ and the $x\text{-intercept}$ is $2\pm \sqrt{6}$ . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: $f\left(x\right)=-x^2-4x+4$