08Csolvingbygraph

= Solving Quadratic Equations with Graphs =

** Recall **
 * The graph of a quadratic is called a parabola
 * For any graph, the ** x-intercepts ** occur where ** y = 0 **

Hence,

the solution to ax 2 + bx + c = 0 can be obtained by
 * sketching y = ax 2 + bx + c
 * Locating the x-intercepts (where y = 0)

** Example 1 **
Use a graph to solve _ _ 2x 2 – 4x – 6 = 0


 * Solution: **

Sketch y = 2x 2 – 4x – 6

X-Intercepts
 * y = 0 when
 * ** x = –1 ** ... and ... ** x = 3 **

Hence _ _ solutions to 2x 2 – 4x – 6 = 0 _ _ are: ... ** x = –1 ** ... and ... ** x = 3 **


 * Confirm Solutions: **

{Substitute x = –1 into original equation}

math \\ . \qquad 2x^2 - 4x - 6 = 0 \qquad \big\{ \text{sub } x = -1 \big\} \\. \\ . \qquad 2(-1)^2 - 4(-1) - 6 = 0 \\. \\ . \qquad \quad 2 + 4 - 6 = 0 \\. \\ . \qquad \qquad 0 = 0 math

{substitute x = 3 into original equation}

math \\ . \qquad 2x^2 - 4x - 6 = 0 \qquad \big\{ \text{sub } x = 3 \big\} \\. \\ . \qquad 2(3)^2 - 4(3) - 6 = 0 \\. \\ . \qquad \quad 18 - 12 - 6 = 0 \\. \\ . \qquad \qquad 0 = 0 math

... ... ... Where the x-intercepts are not clearly on a grid-mark ... ... ... you will need to give an __**approximate**__ solution.
 * Note: **


 * Recall **
 * some quadratic equations have one solution
 * some quadratic equations have no real solutions


 * Example 2 **

Use a graph to solve:
 * 1) x 2 – 4x + 4 = 0
 * 2) –x 2 – 3x – 3 = 0


 * Solution: **

Sketch the graphs
 * 1) y = x 2 – 4x + 4
 * 2) y = –x 2 – 3x – 3

X-Intercepts

1. ........ y = x 2 – 4x + 4 ... ... ... ... x = 2 when y = 0 hence ... .... .. the solution to x 2 – 4x + 4 = 0 ... is ... ** x = 2 **

2. ... ... y = –x 2 – 3x – 3 ... ... ... ... has no x-intercepts hence ... ... .. –x 2 – 3x – 3 = 0 ... has ... ** no real solutions **


 * Confirming Solutions: **

{substitute x = 2 into equation 1}

math \\ . \qquad x^2 - 4x + 4 = 0 \qquad \big\{ \text{sub } x = 2 \big\} \\. \\ . \qquad (2)^2 - 4(2) + 4 = 0 \\. \\ . \qquad \quad 4 - 8 + 4 = 0 \\. \\ . \qquad \qquad 0 = 0 math

{no solutions to confirm for equation 2}

Graphing with Technology
Most graphing packages include a function to find the x-intercepts of a graph. Most of them will give the decimal approximation rather than an exact answer.


 * [[image:bhs-methods10/08Cgrph4.gif width="391" height="378" align="right"]]Graphmatica **

Draw the graph Go to the ** Calculus ** menu Select ** Find Critical Points **
 * The x-intercepts will appear in a pop-up-box.
 * They will be called "** zeros **" {because they occur where y = 0}
 * You will also see the turning point

On the right we see Hence
 * y = x 2 + x – 4
 * has x-intercepts (zeros) at
 * x = –2.56
 * x = 1.56
 * also the turning point is a __minimum__ at (–0.5, 4.25)
 * the solutions to x 2 + x – 4 = 0 are (approx)
 * x = –2.56
 * x = 1.56


 * [[image:08Cgrph5.GIF width="292" height="358" align="right"]]Casio Classpad **


 * Go to the Graphs/Tables screen
 * Draw the graph
 * Go to ANALYSIS Menu, G-SOLVE Submenu
 * Select "Root" {solutions are sometimes called __**roots**__}
 * The coordinates of __one__ of the x-intercepts will appear at the bottom of the screen. (xc, yc)
 * Use the blue side arrows to switch between the two x-intercepts

{The screenshot at right shows that one of the solutions to ... ... ... x 2 + x – 4 = 0 ... ... ... is (approx) x = –2.56

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