Exercise 25 Page 234

We are asked to solve the given quadratic equation. We will solve it by graphing. There are three steps to solving a quadratic equation by graphing.

1. Write the equation in standard form, $a{x}^2+bx+c=0.$
2. Graph the related function $f(x)=a{x}^2+bx+c.$
3. Find the $x\text{-}$intercepts, if any.

The solutions, or roots, of $a{x}^2+bx+c=0$ are the $x\text{-}$intercepts of the graph of $f(x)=a{x}^2+bx+c.$ Let's write our equation in standard form. This means gathering all of the terms on the left-hand side of the equation. Now we can identify the function related to the equation.

Graphing the Related Function

To draw the graph of the related function written in standard form, we must start by identifying the values of $a,$ $b,$ and $c.$ We can see that $a=2,$ $b=\text{-}8,$ and $c=32.$ Now, we will follow three steps to graph the function.

1. Find the axis of symmetry.
2. Make a table of values using $x$ values around the axis of symmetry.
3. Plot and connect the points with a parabola.

Finding the Axis of Symmetry

The axis of symmetry is a vertical line with equation $x=\text{-}\frac{b}{2a}.$ Since we already know the values of $a$ and $b,$ we can substitute them into the formula. The axis of symmetry of the parabola is the vertical line with equation $x=2.$

Making the Table of Values

Next, we will make a table of values using $x$ values around the axis of symmetry $x=2.$

$x$	$2x^2-8x+32$	$f(x)$
$0$	$2(0{)}^2-8(0)+32$	$32$
$2$	$2(2{)}^2-8(2)+32$	$24$
$4$	$2(4{)}^2-8(4)+32$	$32$

Plotting and Connecting the Points

We can finally draw the graph of the function. Since $a=2,$ which is positive, the parabola will open upwards. Let's connect the points with a smooth curve.

Finding the $x\text{-}$intercepts

Let's identify the $x\text{-}$intercepts of the graph of the related function.

We can see that the parabola does not intersect the $x\text{-}$axis. Therefore, the equation $2x^2-8x=\text{-}32$ has no real solutions.