Chapter 7.1, Problem 49PPSE

Solution

To Find:The graph that represents the equation $y=x^2-x-2$

The graph represents the function is option $\left(G\right)$ .

Given:

  

  

  

  

Explanation:

Given the expression $y=x^2-x-2$

:

To find the graph that represents the expression, find the properties of the function.

Rewrite the equation in vertex form.

Complete the square for $x^2-x-2$ :

Use the form $a{x}^2+bx+c$ , to find the values of $a,b,$ and $c$ :

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

Consider the vertex form of a parabola:

  $a{\left(x+d\right)}^2+e$

Find the value of $d$ using the formula $d=\frac{b}{2a}$ :

  $\begin{array}{c}d=\frac{-1}{2\cdot 1}\\ d=-\frac{1}{2}\end{array}$

Find the value of $e$ using the formula $e=c-\frac{b^2}{4a}$ :

Substitute the values of $c,b$ and $a$ into the above formula:

  $\begin{array}{c}e=-2-\frac{{\left(-1\right)}^2}{4\cdot 1}\\ e=\frac{-2\cdot 4-1}{4}\\ e=\frac{-9}{4}\end{array}$

Substitute the values of $a,d$ and $e$ into the vertex form ${\left(x-\frac{1}{2}\right)}^2-\frac{9}{4}$ :

  ${\left(x-\frac{1}{2}\right)}^2-\frac{9}{4}$

Set $y$ equal to the new right side:

  $y={\left(x-\frac{1}{2}\right)}^2-\frac{9}{4}$

Use the vertex form, $y=a{\left(x-h\right)}^2+k$ , to determine the values of $a,h$ and $k$ :

  $\begin{array}{c}a=1\\ h=\frac{1}{2}\\ k=-\frac{9}{4}\end{array}$

Since the value of $a$ is positive, the parabola opens up.

Find the vertex $\left(h,k\right)$ :

  $\left(\frac{1}{2},-\frac{9}{4}\right)$

Find $p$ , the distance from the vertex to the focus.

Find the distance from the vertex to a focus of the parabola by using the following formula $\frac{1}{4a}$ :

Substitute the value of $a$ into the formula:

  $\begin{array}{l}=\frac{1}{4\cdot 1}\\ =\frac{1}{4}\end{array}$

Find the focus.

The focus of a parabola can be found by adding $p$ to the $y-$ coordinate $k$ if the parabola opens up or down.

  $\left(h,k+p\right)$

Substitute the known values of $h,p,$ and $k$ into the formula and simplify:

  $\left(\frac{1}{2},-2\right)$

Find the axis of symmetry by finding the line that passes through the vertex and the focus:

  $x=\frac{1}{2}$

Find the directrix:

The directrix of a parabola is the horizontal line found by subtracting $p$ from the $y-$ coordinate $k$ of the vertex if the parabola opens up or down:

  $y=k-p$

Substitute the known values of $p$ and $k$ into the formula and simplify:

  $y=-\frac{5}{2}$

Use the properties of the parabola to analyse and graph the parabola:

Select a few  $x$ values, and plug them into the equation to find the corresponding  $y$ values. The  $x$  values should be selected around the vertex.

Replace the variable $x$ with $-2$ in the expression:

  $\begin{array}{c}f\left(-2\right)={\left(-2\right)}^2-\left(-2\right)-2\\ =4+2-2\\ =4\end{array}$

Replace the variable $x$ with $-1$ in the expression:

  $\begin{array}{c}f\left(-1\right)={\left(-1\right)}^2-\left(-1\right)-2\\ =1+1-2\\ =0\end{array}$

Replace the variable $x$ with $\frac{1}{2}$ in the expression:

  $\begin{array}{c}f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)-2\\ =-\frac{9}{4}\end{array}$

Replace the variable $x$ with $1$ in the expression:

  $\begin{array}{c}f\left(1\right)={\left(1\right)}^2-\left(1\right)-2\\ =1-1-2\\ =-2\end{array}$

Replace the variable $x$ with $2$ in the expression:

  $\begin{array}{c}f\left(2\right)={\left(2\right)}^2-\left(2\right)-2\\ =4-2-2\\ =0\end{array}$

Create the table:

$x$	$y$
$-2$	$4$
$-1$	$0$
$\frac{1}{2}$	$-\frac{9}{4}$
$1$	$-2$
$2$	$0$

Direction: Opens Up

Vertex: $\left(\frac{1}{2},-\frac{9}{4}\right)$

Focus: $\left(\frac{1}{2},-2\right)$

Axis of Symmetry: $x=\frac{1}{2}$

Directrix: $y=-\frac{5}{2}$

Plot the points:

  

Thus, the graph represents the function is option $\left(G\right)$ .