Transcript to the animated example

GCSE Maths
Number & Algebra
N 58
Graphical Solutions to Quadratic
Functions
Subject Content Reference: N6.7h
We need to be able to find graphical solutions to quadratic functions . .
Example
Solve the quadratic equation x2 + 2x - 5 = 0 using a graph:
Step 1: Draw up a table of values for y = x2 + 2x - 5
y=
x2
x
-3
-2
-1
0
1
2
3
+ 2x - 5
-2
-5
-6
-5
-2
3
10
Step 2: Draw the graph of y = x2 + 2x - 5 . .
y
Step 3:
Now 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) . .
10
here
5
The approximate solutions of the quadratic
equation x2 + 2x - 5 = 0 are
1.4 and -3.4
-3 -2 -1
-5
0
1 2 3
x
When there are two points of intersection,
there are two solutions to the equation . .
. . and here
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 . .
learn
Exercise 1
1) Solve the quadratic equation x2
x
-3
-2
2) Solve the quadratic equation 2x2
+ 3x - 5 = 0 using a graph:
-1
0
y = x2 + 3x - 5
1
2
3
x
-3
-2
- x - 5 = 0 using a graph:
-1
0
y = 2x2 - x - 5
y
y
x
x
1
2
3
We need to be able to find graphical solutions to more complex quadratic equations . .
Example
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
+ 2x - 5
-2
-5
-6
-5
-2
3
10
x
-3
-2
-1
0
1
2
3
y = 2x + 1
-5
y=
Step 2: Find two values for the linear graph y = 2x + 1 . .
y
x2
1
Step 3: Draw the graphs of y = x2 + 2x - 5 and y = 2x + 1 . .
10
Step 4: Now use the graphs to find approximate solutions by looking at the
points where the graphs intersect . .
here
5
The approximate solutions of the quadratic
equation x2 + 2x - 5 = 2x + 1 are
-3 -2 -1
0
1 2 3
x
2.4 and -2.4
When there are two points of intersection,
there are two solutions to the equation . .
-5
. . and here
Be sure to give them both!
Algebraic check: x2
+ 2x - 5 = 2x + 1 becomes x2 = 6 (taking 2x from and adding 5 to both sides)  x = √6 = ± 2.45
learn
Exercise 2
1) Solve the quadratic equation 2x2
x
-3
-2
+ 3x - 5 = 2x + 3 using graphs:
-1
0
1
2
3
2) Solve the quadratic equation 2x2
x
y = 2x2 + 3x - 5
y = 2x2 - 3x + 1
y = 2x + 3
y = 2x - 3
y
-3
-2
- 3x +1 = 2x - 3 using graphs:
-1
0
y
x
x
1
2
3
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)
2) Find the linear equation to use with the quadratic graph
answer
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  x2 - 2x - mx + 1 - c = 0 (subtracting mx + c from both sides)
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
(because -2x - - 5x = 3x)
(because 1 - 2 = -1)
So the linear equation we need to use is y = -5x + 2
answer
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
y=
x2
-3
-2
-1
0
1
2
3
1)
y
a)
x2 + 3x - 2 = 0
b)
x2 + 3x - 2 = -3
a)
4 - x2 = x2 + 3x - 2
+ 3x - 2
y = x2 - 2x + 3
y = 4 - x2
2)
x
y
x
a)
x2 - 2x + 3 = 5
b)
x2
4)
- 2x + 3 = 12
y
x
b) 4
3)
y
x
a)
4 - x2 = 0
b)
x2 - 2x + 3 = 4 - x2
- x2 = 2x + 1
c) Show, using graphs, there is just one solution to the equation
x2 + 3x - 2 = x2 - 2x + 3