Second Term Lesson Note for Week Three.
Class : SSS 1
Subject : Mathematics
Topic : Graphical Method of quadratic equation.
Duration : 80 Minutes
Period : Double Period
Reference Book :
- New General Mathematics for Senior Secondary School, SSS 1.
- Lagos State Unified schemes of work for Senior Secondary School, SSS 1 – 3.
- Online Resources
Instructional Material : Chart showing the graphs of quadratic Equations.
Learning Objectives : By the end of the lesson the students will be able to :
i. Reading the roots from the graph
ii. Determination of the minimum and maximum values
iii. Line of symmetry.
Content :
Graphical Methods of Quadratic Equation.
The following steps should be taken when using graphical method to solve quadratic equation :
1. Use the given range of values of the independent variable (usually x ) to determine the corresponding values of the dependent variable (usually y ) by the quadratic equation or relation given. If the range of values of the independent variable is not given, choose a suitable one.
2. From the results obtained in step (i), prepare a table of values for the given quadratic expression.
3. Choose a suitable scale to draw your graph.
4. Draw the axes and plot the points.
5. Use a broom or flexible curve to join the points to form a smooth curve.
Notes :
1. The roots of the equation are the points where the curve cuts the x – axis because along the x- axis y = 0
2. The curve can be an inverted ∩ – shaped parabola or it can be a U-shaped parabola. It is n-shaped parabola when the coefficient of x² is negative and it is U- shaped parabola when the coefficient of x² is positive.
Maximum value of y occurs at the peak or highest point of the n-shaped parabola while minimum value of y occurs at the lowest point of V-shaped parabola.
3. The curve of a quadratic equation is usually in one of three positions with respect to the x – axis.
(a) (b) (c)
In fig(a), the curve crosses the x-axis at two clear points. These two points give the roots of the quadratic equation.In fig (b), the two points are coincident, i.e their points are so close together that the curve touches the x axis at one point. This corresponds to an equation which has one repeated root.
In fig (c), the curve does not cut the x-axis. The roots of an equation which gives a curve in such a position are said to be imaginary.
4. The line of symmetry is the line which divides the curve of the quadratic equation into two equal parts.
Examples
1a. Draw the graph of y =11 + 8x – 2x² from x = -2 to x = +6.
b. Hence find the approximate roots of the equation 2x² – 8x – 11= 0
c. From the graph, find the maximum value of y.
2a. Given that y = 4x² – 12x + 9 ,copy and complete the table below :
| x | – 1 | 0 | 1 | 2 | 3 | 4 |
| 4x² | 4 | 16 | 64 | |||
| -12 | 12 | – 24 | – 48 | |||
| +9 | 9 | 9 | 9 | |||
| y | 25 | 1 | 3 | 25 |
b. Hence draw a graph and find the roots of the equation 4x² – 12x + 9 = 0
c. From the graph, what is the minimum value of y ?
d. From the graph, what is the line of symmetry of the curve?
Solutions :
1.) Y = 11 +8x – 2x²
from x = – 2 to x = + 6
When x = – 2
Y =11+8(-2)-2(-2)²
Y = 11 – 16 -2 ( +4)
Y = 11 – 16 – 8
Y = – 5 – 8 = -13.
When x = -1
Y= 11 + 8 (-1) -2 (-1)2
Y= 11 – 8 – 2 ( + 1)
Y = 11 – 8 -2
Y = 3 -2 = 1.
When x = 0
Y = 11 + 8 (0) – 2 (0) 2
Y = 11 + 0 – 2 x 0
Y =11+ 0 – 0
Y=11
When x=1
Y = 11 + 8 ( 1) -2 ( 1)2
Y = 11 + 8 – 2 x 1
Y = 19 -2 = 17
When x =2
Y = 11 + 8 (2) -2 (2)2
= 11 + 16 – 2 x 4
= 27 – 8 = 19
when x = 3
y = 11 + 8 ( 3) – 2 ( 3) 2
= 11 + 24 – 2 x 9
= 35 – 18 = 17
when x = 4
y = 11 + 8 (4) – 2 (4) 2
= 11 + 32 – 2 x 16
= 43 – 32 = 11
when x = 5
y = 11 + 8 (5) -2 ( 5)2
= 11 + 40 -2 x 25
= 51 – 50 = 1
when x = 6
y = 11 + 8 ( 6) – 2 (6)
= 11 + 48 -2 x 36
= 59 – 72
=-13
The table of values is given below :
| X | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Y | -13 | 1 | 11 | 17 | 19 | 17 | 11 | 1 | – 13 |
Scale
On x axis, let 2cm = 1 unit; on y axis, let 1cm = 5 units
b. From the graph, the approximate roots of the equation are the points where the curve cuts the x axis,this is so because
y = 11 + 8x – 2×2
-1 x y = -1 x (11) + 8x ( – 1) – 2×2 (-1)
-y = -11 – 8x + 2×2
-y = 2×2 – 8x – 11 = 0
-1x – y = 0 x -1
i.e y = 0
Thus, from the graph, the roots of the equation 2x² – 8x – 11 = 0 are x = -1.1 or x = 5.1
c. The maximum value of y = 19.
2 a. The completed table is given as follows :
| x | – 1 | 0 | 1 | 2 | 3 | 4 |
| 4x² | 4 | 0 | 4 | 16 | 36 | 64 |
| – 12x | 12 | 0 | – 12 | – 24 | – 36 | – 48 |
| +9 | 9 | 9 | 9 | 9 | 9 | 9 |
| y | 25 | 9 | 1 | 1 | 9 | 25 |
Scale :
On x axis, let 2cm = 1unit and on y-axis, let 1cm = 5 units
From the graph, the roots of the equation is the points where the curve touches the x axis, i.e x = 1.5 twice
c. From the graph, the minimum value of y = 0
d. From the graph, the line of symmetry of the curve is line x = 1.5
EVALUATION :
a. Using a suitable scale, draw the graph of y = x² – 2x from x = – 2 to x = + 4
b. From the graph, find the approximate roots of the equation : x² – 2x = 0
c. What is the minimum value of y ?
d. Find the values of x when y = 7.
Finding an equation from a given graph
In general, if a graph (curve) cuts the x axis, at points a and b, the required equation is obtained from the expression ( x – a) ( x – b ) = 0
Examples :
Find the equation of the graphs in the figures below:
Fig. 1 y
6 –
4 –
2 –
-3 -2 -1 1 2 3 x
-5-
-10-
y
15 –
10 –
5 –
-5 -4 -3 -2 -1 1
-2-
-4-
Solutions :
1. First in figure 1 when y = 0, x = – 2 and x = ½
Hence,
x – (-2) x – ½ = 0
x + 2 x – ½ = 0
x ( x – ½) + 2 (x – ½ ) = 0
x² – 1/2x + 2x – 1 =0
x² + 1 ½ x – 1 = 0 ………… ( 1)
Second :
At the intercept in y axis
y = – 2 when x = 0
However, the constant term in equation (1) is – 1
Then, multiply both sides of the equation (1) by 2
i.e 2x² + 3x – 2 = 0 ……………2
Equation (2) satisfies
x -1/2 x – (-2) = 0
and the requirement that the constant term should be -2
:. The equation of the curve is
y = 2x² + 3x – 2 = 0
2. First in fig (2), the curve just touches the x axis at the point x = – 4. Since a quadratic equation has two roots, this implies that the root are repeated when y = 0
i.e when y = 0, x = – 4 (twice )
So the equation must satisfy
(x – (- 4)) (x – (- 4)) = 0
(x + 4) (x + 4) =0
x (x + 4) +4 (x + 4) = 0
x² + 4x + 4x + 16 = 0
x² + 8x + 16 = 0 ……….. Eqn. (1)
Second :
At the intercept on y- axis
y = 15 when x = 0
However, the constant term in eqn. (1) is + 16. Then multiply both sides of the eqn. (1) by – 1.
i.e –x² – 8x – 16 = 0 ……….. Eqn. (2)
Thus, eqn. (2) satisfies
( x + 4) ( x + 4) = 0 and the requirement that the constant term should be – 16.
:. The equation of the curve is
y = -x² – 8x – 16.
EVALUATION :
Find the equations of the graph in the figure below:-
4 –
-1 5/2 -1 4
-2
10 –
4 –
-2 5
-1 2
GENERAL EVALUATION :
a. Draw the graph of y = x² + 2x – 2 from x = – 4 to x = + 2.
b. Hence find the approximate roots of the equation x² + 2x – 2 = 0
2. a. Draw the graph of y = x² – 5x + 6 from x = – 5 to x = + 1
b. Hence find the approximate roots of the equation x² – 5x + 6 = 0
ASSIGNMENT :
Read and study :
New General mathematics SS 1 pages 69 – 74 by MF Macrae et al.
WEEKEND ASSIGNMENT
Use the graph below to answer question 1- 5
-3 -2 -1 1 2 3 4
-2 –
-4 –
-6 –
Find the equation of the graph above
What are the solutions of the equations obtained in question (I) above?
What is the minimum value of y?
From the graph, what is the value of x when y = 2?
From the graph, what is the value of y when x = 1 ½ ?
THEORY
Prepare a table of values for the graph of y = x² + 3x – 4 for values of x from – 6 to + 3
Use a scale of 1cm to 1 unit on both axes and draw the graph.
Find the least value of y
What are the roots of the equation :
x² + 3x – 4 = 0?
Find the values of x, when y = 1