Chapter 3.1, Problem 11E

a.

To determine

To Express: The quadratic function in standard form.

Answer to Problem 11E

the quadratic function is expressed in standard form as $f\left(x\right)=2{\left(x+\frac{3}{2}\right)}^2-\frac{9}{2}$

Explanation of Solution

Given: The function is $f\left(x\right)=\left(2x^2+6x\right)$

Calculation:

The quadratic function $f\left(x\right)=\left(2x^2+6x\right)$ 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 upwards if $a>0$ .

Solve the function:

  $\begin{array}{l}f\left(x\right)=\left(2x^2+6x\right)\\ f\left(x\right)=2\left(x^2+3x\right)\text{ [factor 2 from the }x\text{ terms]}\\ f\left(x\right)=2\left(x^2+3x+\frac{9}{4}\right)-2\cdot \frac{9}{4}\text{ [complete the square : add 2 inside parentheses, subtract }2\cdot \frac{9}{4}\text{ outside]}\\ f\left(x\right)=2{\left(x+\frac{3}{2}\right)}^2-\frac{9}{2}\text{ [factor and multiply]}\end{array}$

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 $f\left(x\right)=2{\left(x+\frac{3}{2}\right)}^2-\frac{9}{2}$

b.

To determine

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

Answer to Problem 11E

the vertex is $\left(h,k\right)=\left(-\frac{3}{2},-\frac{9}{2}\right)$ .

  $x\text{-intercept}=0\text{ and }-3$

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

Explanation of Solution

Given: The function is $f\left(x\right)=\left(2x^2+6x\right)$

Calculation:

The quadratic function $f\left(x\right)=\left(2x^2+6x\right)$ is expressed in standard form as:

  $f\left(x\right)=2{\left(x+\frac{3}{2}\right)}^2-\frac{9}{2}$

by completing the square. The graph of the function f is a parabola with vertex $\left(h,k\right)$ , the vertex is $\left(h,k\right)=\left(-\frac{3}{2},-\frac{9}{2}\right)$ .

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

The $x\text{-intercept}$ is given when $y=0$ , so $2x^2+6x=0$ which implies $2x\left(x+3\right)=0$

Therefore, $x\text{-intercept}=0\text{ and }-3$

c.

To determine

To Sketch: The graph of the quadratic function.

Explanation of Solution

Given: The function is $f\left(x\right)=\left(2x^2+6x\right)$

Graph:

The standard form of the function is:

  $f\left(x\right)=2{\left(x+\frac{3}{2}\right)}^2-\frac{9}{2}$

From the standard form it is observed that the graph is a parabola that opens upward and has vertex $\left(-\frac{3}{2},-\frac{9}{2}\right)$ . As an aid to sketching the graph, find the intercepts. The $y\text{-intercept}=f\left(0\right)=0$ and the $x\text{-intercept}=0\text{ and }-3$ . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: $f\left(x\right)=\left(2x^2+6x\right)$