N 58

Graphical Solutions to Quadratic Functions

Subject Content Reference: N6.7h

GCSE Maths Number & Algebra

We need to be able to find graphical solutions to quadratic functions . .

Example

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Solve the quadratic equation x2 + 2x - 5 = 0 using a graph:

Step 1: Draw up a table of values for y = x2 + 2x - 5x -3 -2 -1 0 1 2 3

y = x2 + 2x - 5 -2 -5 -6 -5 -2 3 10

Step 2: Draw the graph of y = x2 + 2x - 5 . .

10

5

0

-5

1 2 3-1-2-3x

yNow use the graph to find approximate solutions by looking at the points where the graph intersects the line y = 0 (i.e. the x-axis) . .

here

. . and here

The approximate solutions of the quadraticequation x2 + 2x - 5 = 0 are

1.4 and -3.4

When there are two points of intersection,there are two solutions to the equation . .

Be sure to give them both!

Remember - a quadratic graph is symmetrical. Here, if we just connect the plotted points,

the graph only intersects the line y = 0 once - but it can easily be extended because of its symmetry . .

Step 3:

Exercise 1

x

y

x -3 -2 -1 0 1 2 3

y = x2 + 3x - 5

x

y

x -3 -2 -1 0 1 2 3

y = 2x2 - x - 5

1) Solve the quadratic equation x2 + 3x - 5 = 0 using a graph: 2) Solve the quadratic equation 2x2 - x - 5 = 0 using a graph:

We need to be able to find graphical solutions to more complex quadratic equations . .

Example

learnlearn

Solve the quadratic equation x2 + 2x - 5 = 2x + 1 using graphs:

Step 1: Draw up a table of values for y = x2 + 2x - 5 . .x -3 -2 -1 0 1 2 3

y = x2 + 2x - 5 -2 -5 -6 -5 -2 3 10

Step 3: Draw the graphs of y = x2 + 2x - 5 and y = 2x + 1 . .10

5

0

-5

1 2 3-1-2-3x

y

Now use the graphs to find approximate solutions by looking at the points where the graphs intersect . .

here

. . and here

The approximate solutions of the quadraticequation x2 + 2x - 5 = 2x + 1 are

2.4 and -2.4

When there are two points of intersection,there are two solutions to the equation . .

Be sure to give them both!

Step 4:

Step 2: Find two values for the linear graph y = 2x + 1 . .x -3 -2 -1 0 1 2 3

y = 2x + 1 -5 1

Algebraic check: x2 + 2x - 5 = 2x + 1 becomes x2 = 6 (taking 2x from and adding 5 to both sides) x = √6 = ± 2.45

Exercise 2

x

y

x -3 -2 -1 0 1 2 3

y = 2x2 + 3x - 5

y = 2x + 3

x

y

x -3 -2 -1 0 1 2 3

y = 2x2 - 3x + 1

y = 2x - 3

1) Solve the quadratic equation 2x2 + 3x - 5 = 2x + 3 using graphs: 2) Solve the quadratic equation 2x2 - 3x +1 = 2x - 3 using graphs:

Examples

1) Show that the points where the graphs y = 4x - 3 and y = 5/2x intersect are the solutions to the quadratic equation 8x2 - 6x - 5 = 0:

At the points of intersection of these graphs, 4x - 3 = 5/2x

8x2 - 6x = 5 (multiplying both sides by 2x)

8x2 - 6x - 5 = 0 (subtracting 5 from both sides) answer

2) Find the linear equation to use with the quadratic graph y = x2 - 2x + 1 to solve the equation y = x2 + 3x - 1

Let the linear equation be y = mx + c, and this intersects with y = x2 - 2x + 1

At the points of intersection, x2 - 2x + 1 = mx + c (subtracting mx + c from both sides) x2 - 2x - mx + 1 - c = 0

Comparing this equation with y = x2 + 3x - 1 when y = 0 we have x2 - 2x - mx + 1 - c = x2 + 3x - 1

- 2x - mx = 3x so m = -5

and 1 - c = -1 so c = 2

So the linear equation we need to use is y = -5x + 2 answer

(because -2x - - 5x = 3x)

(because 1 - 2 = -1)

Exercise 3

1) Show that the points where the graphs y = 5x - 2 and y = 3/2x intersect are the solutions to the quadratic equation 10x2 - 4x - 3 = 0:

2) Find the linear equation to use with the quadratic graph y = x2 - 3x + 2 to solve the equation y = x2 + 2x - 4

Exercise 4

Complete the table of values below, draw the graphs and use them to solve the following equations:

x -3 -2 -1 0 1 2 3

y = x2 + 3x - 2

y = x2 - 2x + 3

y = 4 - x2

y

x

a) x2 + 3x - 2 = 0

b) x2 + 3x - 2 = -3

y

x

a) x2 - 2x + 3 = 5

b) x2 - 2x + 3 = 12

1)

2)

y

x

a) 4 - x2 = 0

b) x2 - 2x + 3 = 4 - x2

3)

y

x

a) 4 - x2 = x2 + 3x - 2

b) 4 - x2 = 2x + 1

4)

c) Show, using graphs, there is just one solution to the equation

x2 + 3x - 2 = x2 - 2x + 3