Chapter 3.1, Problem 22E

a.

To determine

To Express: The quadratic function in standard form.

Answer to Problem 22E

the quadratic function is expressed in standard form as $f\left(x\right)=6{\left(x+1\right)}^2-11$

Explanation of Solution

Given: The function is $f\left(x\right)=\left(6x^2+12x-5\right)$

Calculation:

The quadratic function $f\left(x\right)=\left(6x^2+12x-5\right)$ 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:

  $\begin{array}{l}f\left(x\right)=\left(6x^2+12x-5\right)\\ f\left(x\right)=6\left(x^2+2x\right)-5\text{ [factor 6 the }x\text{ terms]}\\ f\left(x\right)=6\left(x^2+2x+1\right)-5-\left(6\cdot 1\right)\text{ [complete the square: add 1 to parentheses, subtract }\left(6\cdot 1\right)\text{ outside]}\\ f\left(x\right)=6{\left(x+1\right)}^2-11\text{ [factor and multiply]}\end{array}$

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 $f\left(x\right)=6{\left(x+1\right)}^2-11$

b.

To determine

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

Answer to Problem 22E

the vertex is $\left(h,k\right)=\left(-1,-11\right)$ .

The $x\text{-intercept}$ is $-1\pm \frac{\sqrt{66}}{2}$

  $y\text{-intercept}=f\left(0\right)=-5$

Explanation of Solution

Given: The function is $f\left(x\right)=\left(6x^2+12x-5\right)$

Calculation:

The quadratic function $f\left(x\right)=\left(6x^2+12x-5\right)$ is expressed in standard form as:

  $f\left(x\right)=6{\left(x+1\right)}^2-11$

by completing the square. The graph of the function f is a parabola with vertex $\left(h,k\right)$ , the vertex is $\left(h,k\right)=\left(-1,-11\right)$ .

From the standard form it is observed that the graph is a parabola that opens upward and has vertex $\left(h,k\right)=\left(-1,-11\right)$ . As an aid to sketching the graph, find the intercepts.

The $y\text{-intercept}=f\left(0\right)=-5$ . To find the $x\text{-intercepts}$ , set $f\left(x\right)=0$ and factor the resulting equation.

Use Quadratic formula:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{Here, }a=6,b=12,c=-5\\ x=\frac{-12\pm \sqrt{{\left(12\right)}^2-4\left(6\right)\left(-5\right)}}{2\left(6\right)}\\ x=\frac{-12\pm \sqrt{144+120}}{12}\\ x=\frac{-12\pm \sqrt{264}}{12}=\frac{-12\pm 2\sqrt{66}}{12}=\frac{-6\pm \sqrt{66}}{12}=-1\pm \frac{\sqrt{66}}{2}\end{array}$

The $x\text{-intercept}$ is $-1\pm \frac{\sqrt{66}}{2}$

c.

To determine

To Sketch: The graph of the quadratic function.

Explanation of Solution

Given: The function is $f\left(x\right)=\left(6x^2+12x-5\right)$

Graph:

The standard form of the function is:

  $f\left(x\right)=6{\left(x+1\right)}^2-11$

From the standard form it is observed that the graph is a parabola that opens upward and has vertex $\left(-1,-11\right)$ . As an aid to sketching the graph, find the intercepts.

The $y\text{-intercept}=f\left(0\right)=-5$ and The $x\text{-intercept}$ is $-1\pm \frac{\sqrt{66}}{2}$ . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: $f\left(x\right)=\left(6x^2+12x-5\right)$