Algabraic Expressions

TEACHERS GUIDE

Subject: MATHEMATICS

TARGET GROUP: S. 2

TOPIC: ALGEBRAIC EXPRESSIONS

SUB-TOPIC: FUNCTIONS OF THE SECOND ORDER

BRIEF DESCRIPTION OF TOPIC:

Quadratic functions are part of algebraic expressions with one of their variables raised to the power of two. The term quadratic comes from an English word “quad” which means ‘four’. But in mathematics quadratic means ‘squared’.

Algebraic expressions in the form: ax² + bx + c are known as quadratic expressions.

When we write the above expression as y = ax² + bx + c or f(x) = ax² + bx + c, we call it a quadratic function.

It is from such a quadratic function that we can draw a quadratic graph.

When ax² + bx + c = 0 (that is, when y = 0), we call it a quadratic equation.

y = ax² + bx + c is also called a function of second order where a is the coefficient of x², b is the coefficient of x and c is a constant. For all the expressions and functions above the constants a, b, and c are real numbers and a  0; if a is zero, the function is no longer quadratic but linear.

So all functions in which the highest power of the variable is two (that is, x²) are said to be quadratic. If x has a higher power than 2, it is no longer quadratic and must be given another name.

The study of these expressions has produced fruit in modern day communication where it has helped in designing giant radio telescopes, satellite TV dishes and integrated circuits for computers, DVD players and mobile phones.

Algebraic expressions are also used to determine the rate of change of velocity of a moving object. This is what we call acceleration. Motorists need to know when to step on the brakes if they are to avoid accidents. These expressions ably show that doubling the speed of a vehicle makes the stopping distance four times longer and vice versa. That is why we need to slow down in urban areas.

But algebraic expressions can also be used in sports, for example, in determining how and when to kick a drop-goal in rugby, how to shoot in basketball and netball and how to loop the goalkeeper in football.

We also use algebraic expressions in studying population change of species, the spread of epidemics and many others. All the complex calculations start with these simple algebraic expressions, and that is why you ought to study them.

By the end of this subtopic, the student should be able to:

• draw quadratic graphs.

Common terms

Ensure that the students become familiar with the following vocabulary:

(i) quadratic graph (ii) quadratic equation (iii) maximum point

(iv) Axis of symmetry (v) vertex/turning point (vi) quadratic function

(vii) parabola (viii) minimum point (ix) roots

By this time the students should be familiar with:

(i) the use of graphs

2. How to draw graphs of linear functions like: y = x + 2, y = 3x, etc.

Drawing quadratic functions

In order to draw a graph of a linear function one needs to follow the following steps:

(1) Make a table of results including x and y

(2) Draw the Cartesian plane

(3) Label the axes

(4) Choose an appropriate scale in case it has not been given

(5) Plot the points on a graph paper using either a cross (x) or a dot (.) but NOT both unless for different functions.

6. join the points

We use the same procedure outlined above for the quadratic graphs as well.

Begin by probing the students to describe the path of a ball kicked upwards, a stone thrown at an angle and a rope used for skipping. Here below are the likely answers.

Case 1: When you throw a stone upwards, it usually takes on the shape

Case 2: When two girls are skipping a rope, the rope moves in both upward and downward directions making the shapes below

ie or

When a boy kicks a ball in the air, the ball follows the path of the shape below.

In all the above cases, when an object is thrown upwards or when it bounces; it forms a path in a shape known as a parabola.

• In general, all graphs of quadratic functions have the general shape of a parabola shown above.

• Graphs of quadratic functions will either have a minimum point or a maximum point but not both. These are found at the turning point (the vertex) of the parabola.

• Graphs of quadratic functions have one axis of symmetry. The axis of symmetry is the line about which the graphs are formed.

The Graph of y = x²

The most basic quadratic function is the function y = x² or f(x) = x². This is obtained when a = 1, b = 0 and c = 0. Here the values of x may be referred to as the object values or domain while those of y [or f(x)] make up the image values or range.

ACTIVITY 1 (when coefficient a = 1)

Draw the graph of y = x² for the domain -6 ≤ x ≤ 6

Solution

Step 1 : Make a table of values of x and y = x²

x	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6
Y=x2	36	25	16	9	4	1	0	1	4	9	16	25	36

Step 2 : Draw and label the Cartesian plane

Step 3 : Choose a suitable scale for the axes

Step 4 : Plot the ordered pairs (x, y) from the table

Step 5 : Join the points with a free hand to make a smooth curve

Step 6 : Name/label the curve as y = x²

Choosing the scale: The axes should be marked in such a way that they cover the domain (i.e. all x  values given) and are equally spaced e.g. the domain ^6 < x < 6 has 13 units. In Figure 1 we may choose 1 cm to represent a unit. However, the range (the y  values) has squares of x. So we may choose 1 cm to represent 4 units.

Figure 1

Remember that the points are joined with a free hand. Never use a ruler to join points of the quadratic or other curved graphs.

What can you say about the graph? The graph opens upwards because y - values are positive, has a vertex as its lowest or minimum point and it is wider than those of y = ax², where a > 1. It also has a minimum value as x = 0.

ACTIVITY 2 (when coefficient a > 1)

Draw the graph of y = 3x² for the values of x between ^-3 and ⁺3

Solution:

The domain is -3 < x < 3

x	-3	-2	-1	0	1	2	3
x²	9	4	1	0	1	4	9
y = 3x²	27	12	3	0	3	12	27

Figure 2

What can you say about the graph? The graph opens upwards because y-values are positive, it has a vertex as its minimum point and it is narrower than that of y = x² because the coefficient a = 3 causes it to become narrow, hence when the coefficient a > 1, the graph becomes narrower.

ACTIVITY 3 (when the coefficient 0 < a < 1)

Complete the table below and use it to draw the graph of y = ½ x²

x	-10	-8	-6	-4	-2	0	2	4	6	8	10
x2	100	64	36	16	4	0	4	16	36	64	100
y = 1x2 2	50	32	18	8	2	0	2	8	18	32	50

Figure 3

The graph is a bit wider than that of y = x² because the coefficient a = ½ causes it to open up. Hence the coefficient lies between 0 and 1 i.e. 0 < a < 1.

Note

• The graphs are drawn with a free hand through all the plotted points, hence making a smooth curve.

• The graphs have the shape of a parabola as earlier mentioned.

• In the tables, there are repeated values of y for every additive inverse of x.

• The point (0, 0) is the lowest or minimum point of the graphs.

This implies that the graphs of the nature y = ax² always pass through the origin. The minimum point of such a quadratic function is also called the vertex or turning point of the graph.

Note that if all the above graphs are folded into two along the y-axis, we remain with only one half of the graph. This means that these graphs are symmetrical about the y axis, which is the axis of symmetry.

The equation of this axis of symmetry is the line x = 0

ACTIVITY 4 (when the coefficient a < 0)

Complete the table below and use it to draw the graph of y =x^-2

x	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6
x2	36	25	16	9	4	1	0	1	4	9	16	25	36
y=x-2	-36	-25	-16	-9	-4	-1	0	-1	-4	-9	-16	-25	-36

Figure 4

Note that this graph opens downwards because the y - values are negative even when the x values are positive. Here the vertex is the highest point on the graph. This applies to all functions in which a < 0 (i.e. a is negative)

Notice that there are repeated values of y

• The graphs have the same shape of a parabola but in the opposite direction.

• Both graphs have the same highest or maximum point of the parabola.

• The vertex/turning point of the graph is (0, 0).

• When the graph is folded into two along the y-axis, it is observed that it divides into two equal parts along the y - axis, which is the axis of symmetry.

In conclusion:

• When the coefficient of x² is positive, the graph of the quadratic function has a minimum point and the parabola opens upward ‘’

• When the coefficient of x² is negative, the graph of the quadratic function has a maximum point and the parabola opens downwards ‘’.

• The maximum or minimum point of a parabola is called the vertex/turning point of the parabola.

SUB-TOPIC: GRAPHS OF HARDER QUADRATIC FUNCTIONS

BRIEF DESCRIPTION OF TOPIC:

These are the functions in which the coefficient of b and the constant c may or may not be zero. The presence of these values causes the resulting graphs to behave differently from the ones we just looked at.

MAIN CONTENT AND CONCEPTS TO EMPHASISE

By the end of this subtopic, the student should be able to:

• use the graphs to find the roots/solutions of a quadratic equation;

• find the coordinates of the minimum and maximum points of the graph;

• find the equation of the axis of symmetry.

Common terms

Ensure that the students become familiar with the following vocabulary:

(I) quadratic equation (iii) maximum point

(iv) Axis of symmetry (v) vertex/turning point

(viii) Minimum point (ix) roots

The graph of y = x2 + c

What happens when a constant is added to the basic function y = x²? Let us investigate.

ACTIVITY 5

Show on the graph the position of c using the hint below.

Hint: The constant c that we have just added is the y-intercept is the value of the y-coordinate when the x-coordinate is zero. So we expect c to lie on the y-axis or the line x = 0.

Case 1: When c is positive

When c > 0, the graph shifts c units upwards. The vertex will be at point (0, c)

Case 2: when c is negative

When c < 0 then the graph shifts c-units downwards. The vertex will be at the point (0, ^c).

Case 3: when c is positive and the x-value is negative

For y = ^x² + c the graph will look as here below:

Case 4: when both c and the x-value are negative

Here the function y = ^x²  c will yield a graph of the form

The graphs of perfect squares: y = (x + b)²

Now, what happens if the x - values in the perfect square y = x² were altered by a constant b? Would the graph change? From the basic function y = x², the x - value may be altered either by adding or subtracting a constant b e.g. y = (x + b)² or y = (x  b)².

Note that y = (bx) ² and y = () ² still give us the results we saw above. When b < 0 i.e. y = (x - b) ² the graph shifts b units to the right (or positive direction).

What happens when b is positive i.e. b > 0? Show that the graphs are symmetrical about x = ^b.

If you recall well, you will realise that y = (x + b)² is actually equal to y = x² + 2bx + b² and y = (x  b)² is the same as y = x²  2bx + b². These two functions are also quadratic functions of the form y = ax² + bx + c.

This clearly shows that the graphs of these functions have their vertices either to the left or right of the y-axis (or the line x = 0). If only we could express all quadratic functions in the two forms above, we would be able to determine the x-value of the vertex straight away. You will realise that all perfect squares have a minimum value of zero. This is because perfect squares can never be negative. So the graph of the function y = x² turns upwards when x = 0, and that of y = (x  b)² turns upwards when x - b = 0 i.e. when x = b.

Can you guess where the vertex of the graph of y = (x + 4)² will be?

NB: When y = 0, (x + b)² = 0

 x + b = 0

 x = ^b

So the square should not bother you because it easily disappears when we find the square root of both sides (after all, the square root of zero is still zero).

ACTIVITY 6 (when a = 1, y = x² + bx + c)

(i) Draw the graph of y = x² + 2x – 3 for ^5 < x < 3

(ii) Find the turning point of the graph (iii) Equation of the axis of symmetry.

Since the quadratic function has more terms, then each of the terms is separately included in the table as shown below:

x	-5	-4	-3	-2	-1	0	1	2	3
X2	25	16	9	4	1	0	1	4	9
2x	-10	-8	-6	-4	-2	0	2	4	6
-3	-3	-3	-3	-3	-3	-3	-3	-3	-3
y	12	5	0	-3	-4	-3	0	5	12

Figure 5

From the function y = ax² + bx + c, notice that the vertex has shifted by units to the left i.e. , where b = 2, a = 1. Do you remember why this is so?

So gives = ^1

(b) (i) (^1 , ^4) is the turning point (iii) The equation of the axis of symmetry is the line x = ^1

ACTIVITY 7 (when a = -1, y = -x² + bx + c)

Draw the graph of y = x² + 3x + 10 for the domain -3 < x < 5

(b) Write down the range for the graph

x	-3	-2	-1	0	1	2	3	4	5
X2	9	4	1	0	1	4	9	16	25
x-2	-9	-4	-1	0	-1	-4	-9	-16	-25
+3x	-9	-6	3	0	3	6	9	12	15
+10	+10	+10	+10	+10	+10	+10	+10	+10	+10
y	-8	0	6	10	12	12	10	6	0

Figure 6

Remember that range refers to the image or y - values and domain to the object or x – values for the function. So the range is –8, 0, 6, 10, 12, 12, 10, 6, 0,

ACTIVITY 8: Finding the vertex of the graph

We have already seen that the x - value of the vertex is easy to obtain if the quadratic function is a perfect square. For instance, y = a(x – b)² gives b as the x - coordinate of the vertex. This is because when y = 0, a(x – b)² = 0  x = b .

But what if the function given is not a perfect square? Can we still solve it this way? Yes, we can, but we have to turn part of the function into a perfect square first. Here is how we do it.

Say we have the function y = x² + 2x – 3

Step 1 : Take the constant to the other side i.e. y + 3 = x² + 2x

Step 2 : Make sure that the squared term is left without a coefficient by taking it as a common factor i.e. y + 3 = a(x + 2x). In this case, a = 1,

so y + 3 = x² + 2x

Step 3 : Using brackets, introduce a space for a new term on the left and write outside these brackets the coefficient of x² we got in step 2 above,

Say y + 3 = a ( ) = a(x² + ). Here a = 1, so y + 3 + ( ) = x² + 2x

Step 4 : Take the coefficient of the x-term, divide it by 2 and then square it. Then add it to both sides by placing it inside the brackets on both sides i.e. x-terms is 2, divide by 2 gives 1, square gives 1² = 1 and adding it to both sides gives

y + 3 + 1 = x² + 2x + 1

What you obtain on the right is a perfect square and what is on the left is the y-coordinate of the vertex. Thus y + 4 = (x + 1)^2, where x = ^1 and y = ^4

Therefore the vertex is at point (^–1, ^–4).

Note\ i) x² + 2x + 1 = (x + 1)(x + 1) or (x + 1)². A simple rule for writing perfect squares is: take square root of first term (x²) and add it to square root of last term (1²), then square the sum i.e. (x + 1)^2.

2. The vertex coordinates of a quadratic equation are given by the function y = a(x - h)² + k, where (h, k) is the vertex. Notice that h = .

We have seen from above how to obtain this and that a is still the coefficient of the squared term in x, when a is positive, the parabola is u – shaped and when a is negative, the parabola is -shaped, (inverted U-shape).

ACTIVITY 9: Find the vertex of the function y = 3x² + 4x + 2

Step 1: Take the constant to the other side i.e. y – 2 = 3x² + 4 x

Step 2: Factorise out coefficient of squared term i.e. y – 2 = 3(x² + )

Step 3: Introduce a space for new term on the left hand side with coefficient of squared term present in front of these brackets i.e.

y – 2 + 3 ( ) = 3(x² + + )

Step 4: Take the coefficient of the x-term divide it by 2, square it and then place it inside the brackets on both sides i.e. Coefficient of x term is ; half of this is ; square of this is =

Placing it on both sides gives

y – 2 + 3 = 3(x² + + )

 y – 2 + = 3(x + )² or y = 3(x + )² +

Comparing this with our general function y = a(x – h)² + k gives

^–h = and k =

 (h, k) = ( , )

ACTIVITY 10: Graphical solution of y = 0 (and other values of y.)

We saw earlier that a quadratic function y = ax2 + bx + c gives a quadratic equation ax² + bx + c = 0, when y = 0. To solve this equation

(a) Draw the graph of y = x² – 6x + 9 for 0 < x < 3

(b) Use your graph to solve:

(i) x² – 6x + 9 = 0 (ii) x² – 6x + 9 = 2

(iii) x² – 6x + 9 = ^–1

Step 1: find the values of y for x values between 0 and 3 inclusive.

x	0	1	2	3	4	5	6
x2	0	1	4	9	16	25	36
-6x	0	-6	-12	-18	-24	-30	-36
+9	+9	+9	+9	+9	+9	+9	+9
y	9	4	1	0	1	4	9

Step 2: draw the graph using the x and y-values as in Figure 7.

Figure 7

The graphical solution of y = 0 means that we need to find from the graph values of x for which y = 0. So we draw the line y = 0 on the graph as well. Where this line intersects the graph gives the x - value.

So from the function y = x² – 6x + 9 we can say that y = 0 when x² – 6x + 9 = 0 from which we obtain the x - values that satisfy the condition y = 0. We refer to the x - values obtained as the solution for y = 0

Step 3: Draw the line y = 0 on the graph above, then read the x-coordinates where the graph x2 – 6x + 9 = 0 cuts the line y = 0.

From the graph, x2 – 6x + 9 = 0 has only one solution; it is x = 3

(ii) For x² – 6x + 9 = 2;

- Draw and label the line y = 2 on the graph

- Then read the x-coordinates where the graph meets the line y = 2

For x² – 6x + 9 = 2 the graphical solution for y = 2 is

either x = 1.5 or x = 4.5

(iii) x² – 6x + 9 = ^–1. Draw and label the line y = ^1

The line y = ^–1 does not intersect with the graph; there is no solution/roots we can read from the graph for y = ^–1 .

Worked example

(a) Draw the graph of y = x² – x – 6 for ^–4 < x < 3

(b) Use your graph to solve: (i) x² – x – 6 = 0

(ii) x² – x – 6 = ^–4 (iii) x² – x – 6 = 10

( c) Find the (i) minimum point of the graph (ii) equation of axis of symmetry

Solution

x	-4	-3	-2	-1	0	1	2	3	4
x2	16	9	4	1	0	1	4	9	16
-x	4	3	2	1	0	-1	-2	-3	-4
-6	-6	-6	-6	-6	-6	-6	-6	-6	-6
y	14	6	0	-4	-6	-6	-4	0	6

Figure 8

For x² – x – 6 = 0  y = 0

From the graph; the points of intersection of y = x² – x ^– 6 and y = 0 are (^–2, 0) and (3, 0). So either x = ^–2 or x = 3.

(i) For x² – x – 6 = ^–4  y = ^–4 when x² – x – 6 = ^–4

From the graph, there are two points where the line y = ^–4 meets the graph y = x² – x – 6 i.e. (^–1, ^–4) and (2,^–4). So either x = ^–1 or x = 2.

(ii) For x² – x – 6 = 10  y = 10 when x² – x – 6 = 10. Draw and label the line y = 10 on the graph. From the graph, notice that the line y = 10 only meets the graph at one point. i.e. (^–3.5, 10). Thus x = ^–3.5 is the only x - coordinate from the graph

(c ) (i) Minimum point is (0.5, ^–6.5)

(ii) x = 0.5

Model question

(a) Draw the graph of y = x² – 8x + 7 with the domain 0 < x < 7

(b) Use your graph to solve:

(i) x² – 8x + 7 = 0

(ii) x² – 8x + 5 = ^–4

(iii) x² – 8x = ^–12

Solution

x	0	1	2	3	4	5	6	7
x2	0	1	4	9	16	25	36	49
-8x	0	-8	-16	-24	-32	-40	-48	-56
+7	+7	+7	+7	+7	+7	+7	+7	+7
y	7	0	-5	-8	-9	-8	-5	0

Figure 9

(i) x = 1 or x = 7

(ii) x² – 8x + 5 = ^–4

When you look at the left hand side (LHS) you realize that we have x² – 8x + 5 instead of x² – 8x + 7 so we have to adjust it without changing the given question by adding 2 to both sides in order to make the equations similar

i.e. x² – 8x + 5 + 2 = ^–4 + 2

 x² – 8x + 7 = ^–2. This gives y = ^–2

Now draw and label the line y = ^–2 on the graph

 either x = 1.3 or x = 6.6

(iii) x² - 8x = ^–12. Notice that the expression on the left is incomplete.

We need to add 7 to both sides so that we obtain the equation we are solving.

x² – 8x + 7 = -12 + 7

x² – 8x + 7 = -5

Now, draw and label the line y = -5  either x = 2 or x = 6 from the graph.

SCHEME OF WORK

SUBJECT: MATHEMATICS

SCHOOL: YEAR: ………………. TERM: …………..

TOPIC: ALGEBRAIC EXPRESSIONS CLASS: S. 2 No. OF STUDENTS: ……….

NUMBER OF PERIODS PER WEEK: 6

Week	No. of periods	Topic	Subtopic	Aims/objectives	Teaching methods	Teaching aids	References	Remarks
(dates)	4	Algebraic Expressions	Functions of the second order	By the end of this subtopic, the student should be able to: •draw quadratic graphs;	Guided discovery, Demonstrations, Question and answer	chalkboard, graph paper, mathematical/geometry set instruments and manila paper.	MKSM Bk2 SMU Bk2 SSM Bk2
(dates)	6	Algebraic expressions	Graphs of harder quadratic functions	•use the graphs to find the roots/solutions of a quadratic equation; •find the coordinates of the minimum and maximum points of the graph; •find the equation of the axis of symmetry.	Guided discovery, Demonstrations, Question and answer	chalkboard, graph paper, mathematical/geometry set instruments and manila paper.	MKSM Bk2 SMU Bk2 SSM Bk2

LESSON PLAN

Date: …………… Class: S. 3 Period: 1 Number of students: …...

TOPIC: ALGEBRAIC EXPRESSIONS

SUBTOPIC: FUNCTIONS OF THE SECOND ORDER

Lesson objectives

By the end of the lesson the student should be able to:

• Draw quadratic graphs.
• Describe shapes of graphs

Teaching methods:

(i) Guided discovery; (ii) demonstrations (iii) question and answer

Teaching aids: chalkboard, graph paper, mathematical/geometry set instruments and manila paper.

References: MK Secondary Mathematics Bk 2, SSM Bk2, SMU Bk 2

STEPS	TIME	CONTENT	TEACHERS’ ACTIVITY	STUDENTS’ ACTIVITY
I	5 minutes	Review graphs of linear functions and the Cartesian plane. Emphasise notion of ordered pairs	Asks students Explains how to plot x and y coordinates and draw linear graphs	Do as instructed
II	10 minutes	Introduction of the functions of second order and related terms	Explains the meaning of the different terms	Listen attentively and then take notes
III	20 minutes	Description of the general form of quadratic graphs	Asks questions about the form of paths taken by objects thrown in the air	Respond with demonstration or sketches on the chalkboard
IV	20 minutes	Stimulus activity 1: graph of y = x² when a = 1	Asks students to find values of y for the given values of x. Moves around to correct students	Do as instructed in their exercise books.
V	20 minutes	Plotting the graph	Asks students to plot the graph using values obtained above. Moves around to correct students	Plot ordered pairs on graph papers and draw graph
VI	5 minutes	Conclusion and exercise	Recaps the topic and gives exercise	Attempt exercise

Black Board Plan

Date	Topic
	Notes	New Words

Self-Evaluation:……………………………………………………………….

ALGEBRAIC EXPRESSIONS

TIMED ACTIVITY (10 minutes)

1. In a group of eight learners, choose a chairperson and a time keeper.
2. You are to work as a team to come up with a neatly drawn curve y = x² – x -6 in less than ten minutes.
3. The materials required include: ruled paper, squared paper, pencil, pen, ruler, rubber,
4. Complete the table below.

y=x²-x-6

x	-4	-3	-2	-1	0	1	2	3	4	5
x2		9
-x			2
-6
y					-6

5. Agree n a suitable scale and plot the graph y=x²-x-6. Remember to label the curve and the axes. Indicate the scale used on both axes.
6. Write the names of the group members behind the graph paper and hand in if you are satisfied with your work.

(Let the teacher receive the work by piling the sheets in order of handing in to note which group used the least time, teacher should accept work of groups that exceed ten minutes).

7. Work is to be displayed to find which groups presented the best work.
8. How do you rate your work as a group.

The best	V. good	Good	Fair	Lacking

Be honest to yourself.