Patterns and Relationships

Over the past two years, you have developed a strong foundation in core algebraic concepts. In SHS 1, you explored number patterns, algebraic expressions, factorisation, linear equations, and simultaneous equations using both elimination and substitution methods. You also examined functions, graphs, gradients, and line segments. These are skills crucial for analytical thinking.

In SHS 2, you deepened your understanding through the graphical solution of linear equations and practical applications of simultaneous equations. You engaged with sequences and series, enhancing your ability to model and solve real-life problems mathematically.

This year you will review solving quadratic equations using the factorisation method, difference of two squares and the general formula. You will also explore how to solve quadratic equations using graphical methods. Finally, you will interpret quadratic functions, identify maximum and minimum values, equations of the axis of symmetry and integrate both linear and quadratic equations graphically. This will prepare you for advanced problem-solving and real-world mathematical applications.

KEY IDEAS

• Identifying and solving quadratic equations.

• Solving quadratic equations graphically.

• Finding maximum and minimum points.

• Finding equations of axes of symmetry.

• Solving linear and quadratic equations together using graphs.

• Applying quadratic equations to real-life problems.

In Year 1, Section 3, you learned how to solve quadratic equations by factorisation, using algebraic tiles, standard identities, and the general formula.

What Is A Quadratic Function And Equation?

A quadratic function is a mathematical expression of the form: y = ax2 + bx + c, where a, b, and c are real numbers, and a ≠ 0. The graph of a quadratic function is a parabola, which opens upwards if a > 0 or downwards if a < 0 as shown below.

Figure 2.1a: Graph of a parabola (a < 0)

Figure 2.1b: Graph of a parabola (a > 0) The following are examples of quadratic functions

• ,

• ,

• When we set a quadratic function equal to zero, we get a quadratic equation ax2 + bx + c = 0 The following are examples of quadratic equations:

•

•

• Every quadratic equation has a maximum of two roots or solutions.

Methods for solving quadratic equations The following methods can solve quadratic equations.

• Factorisation

• Completing the square

• The general quadratic formula

• Graphical method Finding the roots of a quadratic equation by factorisation The roots (or solutions) of the quadratic equation, ax2 + bx + c = 0, are the values of x that satisfy ax2 + bx + c = 0 Here are the steps for solving quadratic equations by factorisation.

1. Identify a, b and c

2. Calculate the product of a and c, that is: a × c = ac

3. Find two numbers whose product is equal to ac and whose sum is equal to b.

4. Rewrite the middle term by replacing the term bx with the two numbers you found in step 3.

5. Group the first two terms and the last two terms, then factor out common factors from each group.

6. Factor out the common binomial factor to obtain the factored form of the quadratic equation.

7. Set each factor equal to zero and solve for x to find the solutions of the quadratic equation.

Go through the following examples to refresh your knowledge on how to solve quadratic equations by factorisation.

Example 2.1

Find the roots of the equation 3 x 2 +10 x − 8 = 0

Solution

To find the roots of the equation 3x² +10 x − 8 = 0, follow these steps.

1. Multiply the coefficient of ‘x2’ (the term) by the constant term ‘-8’ (the c term):

That is 3(-8) = −24

2. Find two factors that have the product of –24 and the sum of 10 (that is the coefficient of x, or the term).

3. The numbers are: 12 and −2

4. Replace 10 x with 12x and −2x:

3x² +12x −2x − 8 = 0 = 0

5. Group the pair of terms on the left side of the equation and the right-hand side:

(3x²+ 12x) − (2x + 8) = 0

Note: The brackets in (2x + 8) has changed to a positive sign because of the negative in front of 2x just before the bracket.

6. Factor each group: 3x(x + 4) − 2(x + 4) = 0

7. Factor the common binomial: (x + 4)(3x − 2) = 0

8. Equate each factor to zero and solve for :

(x + 4) = 0 or (3 x − 2) = 0 x = -4 or x =

9. Therefore, the roots of the quadratic equation:

3 x² +10x − 8 = 0 are x = −4 or x =

Example 2.2

Find the roots of the equation 2x² − x − 6 = 0

Solution

To find the roots of the equation 2x² − x − 6 = 0, replicate the steps in Example 2.1 as follows.

or or Therefore, the roots of the quadratic equation, are or

Example 2.3

Find the zeros of the function, y = x² − 5x + 6

Solution

To find the zeros of the function, y = x² − 5x + 6 apply the steps in Example 2.1 and solve as follows:

Note: The brackets in (2x − 6) have changed to a negative sign because of the negative multiplier in front of the bracket.

Therefore, the zeros of the function, are

Example 2.4

Find the zeros of the function f(x) = 2x² + 7x + 3

Solution

To find the zeros of the function, f(x) = 2x² + 7x + 3 equate f(x) to zero and follow the steps below.

Therefore, the zeros of the function, f(x) = 2x² + 7x + 3 are

Example 2.5

Solve the equation: x² − 7x + 12 = 0

Solution

Given, x² − 7x + 12 = 0, solve as follows:

Therefore, the values of x from the equation, x² − 7x + 12 = 0, are 3 and 4

Example 2.6

Solve the equation: −x² − 7x − 12 = 0

Solution

Given the equation, −x² − 7x − 12 = 0.

Therefore, the values of from the equation, −x² − 7x − 12 = 0 are −3 and 4.

Solving quadratic equations using the quadratic formula You can also use the quadratic formula to solve any quadratic equation. The formula is derived from completing the square of the general quadratic equation ax2 + bx+c = 0.

You do not need to learn how the formula is derived, but it is helpful to understand where it comes from. What is most important is that you know and can use the quadratic formula correctly.

Given the general quadratic equation: ax2 + bx+c = 0, follow these steps to appreciate where the formula is coming from:

1. Subtract ‘c’ from both sides: ax2 + bx+c − c = 0 − c.

ax2 + bx = −c.

2. Divide both sides of the equation by ‘a’:

3. Divide the coefficient of x by 2:

4. Square and add it to either side of the equation : [ note: ]

5. Complete the squares and simplify the right-hand side:

=

6. Find the square root of both sides:

== ==

7. Subtract from both sides:

=

8. Therefore the roots of the equation, ax2 + bx+c = 0 is x = or x = The general quadratic equation formula is Now, use the general quadratic formula to solve the following questions.

Example 2.7

Solve the quadratic equation, x² + 3x − 10 = 0

Solution

Observe the following steps.

1. Comparing x² + 3x − 10 = 0 with the general quadratic equation:

ax²+ bx + c = 0; a = 1, b = 3 and c = −10

2. Substitute the values of a, b, and c into the general formula:

3. Simplifying:

or or or Therefore, x² + 3x − 10 = 0 is x = 2 or x = −5

Example 2.8

Solve the quadratic equation, 3x² − 2x − 7 = 0

Solution

1. Compare 3x² − 2x − 7 = 0 with the general quadratic equation ax²+ bx + c = 0; a = 3, b = −2 and c = −7

3. Substitute the values of a, b, and c into the general formula:

4. Simplifying:

Therefore, or

Example 2.9

Solve −2x² + 4x + 6 = 0

Solution

1. Compare −2x² + 4x + 6 = 0 with the general quadratic equation:

ax²+ bx + c = 0; a = −2, b = 4 and c = 6

2. Substitute the values of a, b, and c into the general formula:

3. Simplifying:

or Therefore, the values of x for the equation, −2x² + 4x + 6 = 0 are x = −1 or x = 3

Example 2.10

Solve −x² + 3x + 4 = 0

Solution

1. Comparing −x² + 3x + 4 = 0 with the general quadratic equation:

ax²+ bx + c = 0; a = −1, b = 3 and c = 4

2. Substitute the values of into the general formula:

3. Simplifying:

Therefore, the values of x for the equation, −x² + 3x + 4 = 0 are x = −1 or x = 4

Activity 2.1 Identifying and solving quadratic equations Materials Needed

• Graph book/Graph sheet

• Ruler

• Pencil

• Calculators

1. Given the function, y = 2x² − 3x − 1, complete the table below with the given interval, −1 ≤ x ≤ 4 x −1 0 1 2 3 4 y 4 1 19 Using a scale of 2cm to 0.5 units on the x-axis and 2cm to 2 units on the y-axis:

2. Plot the points from the completed table on a graph sheet or graph book.

3. Join all the points with freehand, making a smooth curve.

4. Replace the coefficient of x2 with −2, so we are sketching y = −2x² − 3x − 1

5. Create a table using the new function (y = −2x − 3x − 1) and the given domain, −1 ≤ x ≤ 4.

6. Draw the graph of the function.

7. Compare the two graphs y = 2x − 3x − 1 and y = −2x − 3x − 1.

8. What is your observation? Discuss with a classmate.

9. What effect did changing the sign of the coefficient of 2 have on the shape and direction of the graph?

Activity 2.2 Exploring why a ≠ 0 in a quadratic equation

1. Given the function, y = −3x² − 5x − 1, complete the table below with the given interval, −2 ≤ x ≤ 2 x −2 −1 0 1 2 y −3 −1 −23 Using a scale of 2cm to 0.5 units on the x-axis and 2cm to 4 units on the y-axis:

2. Plot the points on a graph sheet or graph paper.

3. Join all the points with freehand, making a smooth curve.

4. Replace the coefficient of x2 with 0.

5. Write down the new equation.

6. Create a table of values using the function, y = 0x² − 5x − 1 and the given domain, −2 ≤ x ≤ 2.

7. Draw the graph of the function.

8. Compare the two graphs of y = −3x² − 5x − 1 and y = 0x² − 5x − 1.

9. What is your observation? Discuss with a classmate.

10. What happened to the graphs when the coefficient of x2 was replaced with 0

11. How does the presence or absence of the x2 term affect the curvature of the graph?

12. In what ways do the graphs of quadratic functions differ from those of linear functions?

13. Can you now predict the shape of a graph just by looking at the equation?

14. Why is it useful to explore different versions of the same equation by changing coefficients?

15. How can these observations help you in solving real-world problems or designing solutions? Discuss with your classmates.

Real-Life Applications of Quadratic Functions and

Equations Take a look at the following diagram in Figure 2.2.

Figure 2.2: Illustration of a basketball shot path creating a parabola. Source: OIP.jwp-0KMvxIvbyO4Ac- c9MPAAAAA (404×280) Now look at the bridge in Figure 2.3 below.

Figure 2.3: Adomi Bridge showing parabolic shape. Source: maxresdefault.jpg (1280×720) As you can see, both can be explained and studied using quadratic functions and equations.

A quadratic function shows a special kind of relationship between two variables that makes a curve called a parabola. A quadratic equation helps us find the exact points where the curve meets the horizontal axis, or the ground, in real life. Here are some examples of how quadratic equations can help us in real life:

• Solve problems about curved paths, finding the highest or lowest points, and many other real-life situations.

• Indicate precision in sporting activities (for example, when throwing a basketball).

• Plan the best shape for bridges and buildings.

• Predict how things move (like water flowing or cars braking).

• Work with patterns in farming, business, and even fashion.

Study the following real-life examples.

Example 2.11

A rectangular garden in Kumbungu Senior High School has an area of 60m². The length is 4 metres more than the width. Find the dimensions of the garden.

Solution

1. Let the width be x metres. Then the length will be x+ 4 metres.

2. Set up the equation:

x(x + 4) = 60

3. Solve:

x² +4x − 60 = 0 (x + 10) (x − 6) = 0 Either x = −10 or x = 6 Distances cannot be negative; therefore, must be 6, so the dimensions are 6 metres and 10 metres for width and length respectively.

In many Ghanaian communities, sports like football or athletics are common. When a ball is kicked (as in Figure 2.4), its path follows a parabola.

Figure 2.4: Illustration of football shot paths creating a parabola

Example 2.12

Consider the table of values for the equation y = −x² + 5x which represents a basketball’s path, where is the height in metres and is the horizontal distance in metres.

x 0 1 2 3 4 5 y 0 4 6 6 4 0

1. Plot the points and join them smoothly to form the parabola.

Figure 2.5: Graph of y = −x² + 5x

5. Draw the line: y = 3 on the same graph sheet.

This represents the height of the basketball hoop.

Figure 2.6: Graphs of y = −x² + 5x and y = 3

6. Where does the ball’s path meet the hoop?

The ball meets the hoop when −x² + 5x = 3 This implies −x² + 5x − 3 = 0 Substitute a = −1, b = 5 and c = −3 xx = −bb ± √bb²− 4aaaa 2aa Substitute aa = −1, bb = 5 aaaaaa aa = −3 xx = −5 ± √(5)²− 4(−1)(−3) 2(−1) xx = −5 ± √25 − 12 −2 xx = −5 ± √13 −2 xx = −5+√13 −2 or xx = −5−√13 −2 Therefore, the ball’s path meets the hoop at xx = −5+√13 −2 or xx = −5−√13 −2

7. What shape does the graph of a quadratic equation make?

The graph makes a parabola.

8. Why does a basketball shot follow a curved path instead of a straight one?

• Because of gravity.

• When a ball is kicked, it has both horizontal and vertical motions. Gravity pulls it downwards, creating a curved (parabolic) path.

• This type of motion is called projectile motion, modeled by a quadratic equation.

9. How can we find the highest point of the ball’s path?

Use the vertex formula for a parabola:

−bb 2aa, where aa = −1 aaaaaa bb = 5 Substituting into the formula, we have;

−5 2(−1) = (−5)/(−2) = 5/2 = 2.5metres.

47 xx = −bb ± √bb²− 4aaaa 2aa Substitute aa = −1, bb = 5 aaaaaa aa = −3 xx = −5 ± √(5)²− 4(−1)(−3) 2(−1) xx = −5 ± √25 − 12 −2 xx = −5 ± √13 −2 xx = −5+√13 −2 or xx = −5−√13 −2 Therefore, the ball’s path meets the hoop at xx = −5+√13 −2 or xx = −5−√13 −2

7. What shape does the graph of a quadratic equation make?

The graph makes a parabola.

8. Why does a basketball shot follow a curved path instead of a straight one?

• Because of gravity.

• When a ball is kicked, it has both horizontal and vertical motions. Gravity pulls it downwards, creating a curved (parabolic) path.

• This type of motion is called projectile motion, modeled by a quadratic equation.

9. How can we find the highest point of the ball’s path?

Use the vertex formula for a parabola:

−bb 2aa, where aa = −1 aaaaaa bb = 5 Substituting into the formula, we have;

−5 2(−1) = (−5)/(−2) = 5/2 = 2.5metres.

7. What shape does the graph of a quadratic equation make? The graph makes a parabola.

8. Why does a basketball shot follow a curved path instead of a straight one?

• Because of gravity.

• When a ball is kicked, it has both horizontal and vertical motions. Gravity pulls it downwards, creating a curved (parabolic) path.

• This type of motion is called projectile motion, modeled by a quadratic equation.

9. How can we find the highest point of the ball’s path?

Use the vertex formula for a parabola:

, where a = −1 and b = 5 Substituting into the formula, we have; = = = 2.5metres.

Therefore, the highest point of the ball’s path is x = 2.5m when and the height at this point is 6.25m

10. At what horizontal distance(s) did the ball reach the hoop height?

The ball reaches three metres high at or Solving Quadratic Equations Using Graphs A graphical method can also be used to solve quadratic functions and equations. This method helps you get approximate answers to quadratic equations.

How to draw a quadratic graph given the domain

1. Find -values - Plug each -value into the equation to get y.

2. Plot the points (x, y) on the graph.

3. Draw the curve - Sketch a smooth U- shaped curve or ∩-shaped curve through the points. Use a ruler for the axes, and freehand for the curve.

Example 2.13

Given the quadratic equation, y = −0.5x² + 3x + 2 and the domain, x = −1,0,1,2,3,4,5,6,7 Use the information to answer the following questions:

1. Find the corresponding y-values from the x-values.

Substitute the values of x into the given function, y = −0.5x² + 3x + 2 For x = −1; y = (-0.5) (−1)2 + 3(-1)+2 = −1.5 For x = 0; y = (−0.5) (0)2 + 3(0)+2 = 2.

Create a table for x and y values as shown below.

x −1 0 1 2 3 4 5 6 7 y −1.5 2 4.5 6 6.5 6 4.5 2 −1.5

2. Using a scale of 1cm to 1 unit, on the x-axis and 1cm to 5 units on the y-axis, plot the points on a graph and sketch the curve.

Draw the graph as shown below:

Figure 2.7: Graph of y = −0.5x² + 3x + 2 From looking at the graph we can see that the roots of the equation (where the graph cuts the x-axis) are approximately x = −0.6 and x = 6.6 Finding the Minimum and Maximum Value of a Quadratic Graph The general quadratic function is in the form y = ax²+ bx + c, where a, b, and c are constants with . The graph’s turning point is called the vertex. If the graph opens upward, the vertex is a minimum point, which happens when a > 0.

For example, y = 2x²− 4x + 1 has a minimum turning point because the 2x2 term is positive. Refer to Figure 2.8 on the next page.

Figure 2.8: Graph of f(x) = 2x² – 4x + 1 On the other hand, if a quadratic graph opens downward, its vertex is a maximum point. This happens when the coefficient of x2 is negative (a< 0).

For example, y = 2x²− 4x + 1 has a maximum turning point because the x2 term is negative. Refer to the diagram in Figure 2.9.

Figure 2.9: Graph of f(x) = 2x² – 4x + 1 Nature of roots of quadratic functions and equations The nature of a quadratic function or equation can be ‘real’, ‘repeated’ or ‘complex’ (no roots). When the graph touches the axis at two different points, it has real (distinct) roots; if it touches the axis at only one point it has repeated roots, and if it does not touch the axis, it has complex (no real) roots. The following diagrams show the various kinds of roots:

Figure 2.10a: real (distinct) roots

Figure 2.10b: repeated (equal) roots

Figure 2.10c: complex (no real) roots

Example 2.14

1. Draw the graph of the function, y = x²− 6x + 8 for the interval 0 ≤ x ≤ 6 using a scale of 1 cm to 1 unit on the x-axis and 1 cm to 5 units on the y-axis.

Given the function, y = x²− 6x + 8 for the interval 0 ≤ x ≤ 6, create a table as shown below.

For x = 0, y = (0)2 – 6(0) + 8 ⇒ y = 8 For x = 1, y = (1)2 – 6(1) + 8 ⇒ y = 3 For x = 2, y = (2)2 – 6(2) + 8 ⇒ y = 0 For x = 3, y = (3)2 – 6(3) + 8 ⇒ y = -1 For x = 4, y = (4)2 – 6(4) + 8 ⇒ y = 0 For x = 5, y = (5)2 – 6(5) + 8 ⇒ y = 3 For x = 6, y = (6)2 – 6(6) + 8 ⇒ y = 8 x 0 1 2 3 4 5 6 y 8 3 0 −1 0 3 8 The graph of the function, y = x²− 6x + 8 is shown below.

Figure 2.11: Graph of y = x² – 6x + 8

2. From the graph, determine the roots of the equation.

The roots of the equation, y = x²− 6x + 8 are 2 and 4, as these are the points when the graphs cuts the x-axis.

3. Determine the minimum value and the point at which it occurs.

The minimum value is −1 and it occurs at (3, −1).

Example 2.15

Copy and complete the table of values below for the relation y = −0.5x²+ 3x + 2 x 0 1 2 3 4 5 6 y 2 6.5 2 Using a scale of 2 cm to 2 units on both x and y-axis. Draw the graph of the relation.

Solution

x 0 1 2 3 4 5 6 y 2 4.5 6 6.5 6 4.5 2

Figure 2.12: Graph of y = −0.5x²+ 3x + 2

Example 2.16

1. Copy and complete the table of values below for the relation y = −x²+ 4x with the interval −1 ≤ x ≤ 5

-1 0 1 2 3 4 5 Using a scale of 1cm to 2 units on x-axis and 1cm to 2 units on y-axis. Draw the graph of the relation.

For x = −1, y = − (−1)2 + 4(−1) ⇒ y = −5 For x = 0, y = − (0)2 + 4(0) ⇒ y = 0 For x = 1, y = − (1)2 + 4(1) ⇒ y = 3 For x = 2, y = − (2)2 + 4(2) ⇒ y = 4 For x = 3, y = − (3)2 + 4(3) ⇒ y = 3 For x = 4, y = −(4)2 + 4(4) ⇒ y = 0 For x = 5, y = −(5)2 + 4(5) ⇒ y = −5 x -1 0 1 2 3 4 5 y -5 0 3 4 3 0 -5 The graph is shown below.

Figure 2.13: Graph of y = −x²+ 4x

2. Use this graph to find the x-values where the graph cuts the x-axis.

The values of x where the graph cuts the x-axis are 0 and 4.

3. Find the greatest value of y and the value of x at which it occurs.

The greatest value of y is 4, and it occurs at x = 2.

4. Find the coordinates of the turning point.

The coordinates are (2, 4).

Example 2.17

The diagram in Figure 2.14 shows the graph of a quadratic function, y = 0.1x²+ 3x − 1.

Figure 2.14: Graph of y = 0.1x²+ 3x - 1

1. Find the minimum value of y = 0.1x²+ 3x − 1.

2. Find the minimum point.

Solution

1. The minimum value of y is −23.5

2. The minimum point is (−15, −23.5)

Example 2.18

Figure 2.15 is the graph of the function f(x) = −0.2x² + 20.

1. Write down the maximum value of f(x).

2. What value of x does the maximum value occurs.

3. What is the maximum point?

Figure 2.15: Graph of y = −0.2x² + 20

Solution

1. The maximum value of f(x) is 20.

2. The maximum value occurs at x = 20.

3. The maximum point is (0, 20) Real-Life Applications of Quadratic Graphs

Example 2.18

During inter-school games in Tamale, a striker kicks a penalty, and the ball follows a parabolic path. The ball’s height is modelled by, h(t) = −6t² + 12t + 1. Where h(t) is the height of the ball (in metres) and is the time in seconds.

After how many seconds does the ball reach the goalkeeper’s height of 2 metres?

Solution

Given, h(t) = −6t² + 12t + 1, where h(t) = 2 ⇒ −6t² + 12t + 1 = 2 Rearranging; −6t² + 12t + 1 − 2 = 0 ⇒ −6t² + 12t − 1 = 0 Solving using the general quadratic formular where a = −6, b = 12 and c = −1 56

Example 2.18

During inter-school games in Tamale, a striker kicks a penalty, and the ball follows a parabolic path.

The ball’s height is modelled by, h(tt) = −6tt²+ 12tt + 1. Where h(tt) is the height of the ball (in metres) and tt is the time in seconds.

After how many seconds does the ball reach the goalkeeper's height of 2 metres?

Solution

Given, h(tt) = −6tt²+ 12tt + 1, where h(t) = 2 -6t²+ 12t + 1 = 2 Rearranging; -6t²+ 12t + 1 – 2 = 0 -6t²+ 12t – 1 = 0 Solving using the general quadratic formular where aa = −6, bb = 12 aaaaaa cc = −1;

t = ⁻ᵇᵇ±√bb2−4aaaa 2aa t = ⁻¹²±√(12)2−4(−6)(−1) 2(−6) t = ⁻¹²±√144−24 −12 t = ⁻¹²±√120 −12 t = ⁶+ √30 6 or t = ⁶− √30 6 tt ≈ 1.91 oooo tt ≈ 0.87 Therefore, the time taken for the ball to reach 2 metres is 0.087 seconds and again at 1.91 seconds when it is on its way down again.

Example 2.19:

Kwabena, a farmer in the Ashanti Region, wants to build a parabolic arch for a greenhouse using bamboo. The shape of the arch can be modelled by the equation, h(xx) = −0.5xx²+ 4xx, where: h(xx) is the height of the arch in metres at any point, xx meters from one end of the base.

If the base of the arch is 8 metres long, what is the maximum height of the arch, and at what point along the base does it occur?

Solution

The graph can be modelled as shown below.

Therefore, the time taken for the ball to reach 2 metres is 0.087 seconds and again at 1.91 seconds when it is on its way down again.

Example 2.19:

Kwabena, a farmer in the Ashanti Region, wants to build a parabolic arch for a greenhouse using bamboo. The shape of the arch can be modelled by the equation, h(x) = −0.5x² + 4x, where: h(x) is the height of the arch in metres at any point, meters from one end of the base.

If the base of the arch is 8 metres long, what is the maximum height of the arch, and at what point along the base does it occur?

Solution

The graph can be modelled as shown below.

Figure 2.16: Graph of y = −0.5x² + 4x From the graph, the maximum height occurs at x = 4 and the maximum height is 8 metres.

Alternatively, using the vertex formula, where a = −6, b = 12 and c = −1:

57

Figure 2.16: Graph of y = -0.5x² + 4x From the graph, the maximum height occurs at xx = 4 and the maximum height is 8 metres.

Alternatively, using the vertex formula, where aa = −0.5 aaaaaa bb = 4:

x = (−bb)/(2aa) = (−4)/2(−0.5) x = (−4)/(−1) = 4 Now, finding the maximum height:

h(x) = -0.5(4)²+ 4(4) h(x) = -8 + 16 h(x) = 8 Therefore, the maximum height of the arc is 8 metres.

Activity 2.3 Exploring quadratic graphs with real-life situations A ball is thrown into the air. Its height (in metres) above the ground at time, xx seconds is given by the equation, h(xx) = −xx²+ 4xx + 5.

1. On a graph sheet or paper, using an appropriate scale on the x and y-axes, label the x-axis from -1 to 5 and the y-axis from -1 to 9.

2. Create a table of values for both x and y, where y = h(x).

3. Draw the graph using the values of x and y.

4. From the graph, determine the maximum height the ball reaches?

5. After how many seconds does the ball reach the maximum height?

6. How would the graph and the motion of the ball change if the initial speed or direction of the throw were different?

Now, finding the maximum height:

h(x) = −0.5(4)2 + 4(4) h(x) = −8 + 16 h(x) = 8 Therefore, the maximum height of the arc is 8 metres.

Activity 2.3 Exploring quadratic graphs with real-life situations A ball is thrown into the air. Its height (in metres) above the ground at time, seconds is given by the equation, h(x) = −x² + 4x + 5.

1. On a graph sheet or paper, using an appropriate scale on the x and y-axes, label the x-axis from −1 to 5 and the y-axis from −1 to 9.

2. Create a table of values for both x and y, where y = h(x).

3. Draw the graph using the values of x and y.

4. From the graph, determine the maximum height the ball reaches?

5. After how many seconds does the ball reach the maximum height?

6. How would the graph and the motion of the ball change if the initial speed or direction of the throw were different?

7. What real-life factors could affect the shape of this parabola?

8. In each of the above steps, discuss the results with your classmates.

Graphs of Parabolic Functions

The graph of a parabolic function of the form f(x) = y = −0.5x² + 4x takes the shape of either of the diagrams below. These types of graphs are symmetrical, with the line of symmetry occurring at .

Figure 2.17: Graphs of parabolic shapes For graphs with a maximum turning point (shaped like this ∩, when a < 0:

The coordinates of the turning point is and the maximum value is The line of symmetry is The graph is increasing when , i.e. the gradient is positive.

The graph is decreasing when , i.e. the gradient is negative.

Figure 2.18: Line of symmetry on a maxima parabola For graphs with a minimum turning point (shaped like this ∪, when a > 0:

The coordinates of the turning point are and the minimum value is .

The line of symmetry is .

The graph is decreasing when , i.e. the gradient is negative.

The graph is increasing when , i.e. the gradient is positive.

Figure 2.19: Line of symmetry on a minima parabola Therefore, for a quadratic equation of the form, f(x) = ax²− bx + c, the maximum or minimum value is .

If the graph is written in the form f(x) = a(x − h)2 + k the minimum or maximum value is k and the line of symmetry is x = h or x − h = 0 and the turning point is at (h, k).

Example 2.20:

Find the line of symmetry and turning point of the quadratic function, f(x) = 2x²− 4x + 1

Solution

The line of symmetry is where a = 2 and b = −4 Substituting; =1 To find the turning point, first find the minimum value which is f ( = f(1) ⇒ f(1) = 2(1)2 − 4(1) + 1 f(1) = 2 – 4 + 1 = −1 Therefore, the turning point = (1, −1).

Example 2.21:

Determine whether the function, f(x) = −x²+ 6x − 5 has a maximum or minimum point, and find its value.

Solution:

Since a = −1, which is < 0, the graph opens downwards, hence, it has a maximum point.

Maximum point = , where x = = = 3 Now, = 4 Therefore, the maximum point is (3, 4) and the maximum value is 4.

Example 2.22:

The graph of f(x) = 3(x − 2)2 − 7 is given in vertex form. Write the:

1. Line of symmetry

2. Turning point

3. State if it has a maximum or minimum value

4. At what interval is the graph

a. increasing?

b. decreasing?

Solution

1. Since a = 3 > 0, the graph opens upwards, it has a minimum point.

Comparing f(x) = 3(x − 2)2 − 7 with f(x)= a(x − h)2 + k ⇒h = 2 and k = −7 Line of symmetry is x = 2.

2. The turning point is (h, k) = (2, −7)

3. It has a minimum value since a > 0.

4.

a. The graph is increasing when x < 2.

b. The graph is decreasing when x > 2.

Example 2.23:

Study the graph below and answer the questions that follow.

Figure 2.20: Graph of a parabola Find:

i. The coordinates of the maximum point,

ii. The coordinates of the x intercept.

iii. The equation of the line of symmetry.

iv. The coordinates of the y intercept

Solution

i. The coordinates of the maximum point is (2, 7).

ii. The coordinates of the x intercept are (0.2, 0) and (3.8, 0).

iii. The equation of the line of symmetry is x = 2.

iv. The coordinates of the y intercept is (0, −1).

Solving Linear and Quadratic Equations Using Graphs

A linear and a quadratic equation can have up to two solutions. When drawn together on a graph, these are where the straight line and the curve meet. They can meet at two points (two solutions), touch at one point (one solution), or not meet at all (no

solution).

Observe the diagram in Figure 2.21.

Figure 2.21: Linear and quadratic graphs

Example 2.24:

Solve the following equations simultaneously using graphs.

y = x + 2 and x² + 2

Solution

Figure 2.22: Graph of x + 2 and x² + 2 The simultaneous equations of a linear and quadratic equations is given by the point or points of intersection of the line and parabola representing the equations.

Therefore, the solutions are (1, 3) and (0, 2)

Example 2.25:

Solve the following simultaneous equations:

y = x + 2 y = x2

Solution

Figure 2.23: Graph of y = x + 2 and y = x2 Therefore, the solutions are when x = 2, y = 4 and when x = −1, y = 1.

Example 2.26:

a. Copy and complete the table of values for the relation, y = x² + 2 for the interval −2 ≤ x ≤ 4 x −2 −1 0 1 2 3 y 6 3

b. Using a scale of 1 cm to 2 units on the x-axis and 1cm to 5 units on y-axis, draw the graph of the relation for y = x² + 2 the interval −2 ≤ x ≤ 3

c. Use the graph to find the solution set of

i. x² + 2 = 0

ii. x² + 2 = x +2

Solution

a. Substitute the values into the equation y = x² + 2 to find the corresponding values of y.

For x = −2, y = (−2)2 + 2 ⇒ y = 4 + 2 = 6 For x = −1, y = (−1)2 + 2 ⇒ y = 1 + 2 = 3 For x = 0, y = (0)2 + 2 ⇒ y = 0 + 2 = 2 For x = 1, y = (1)2 + 2 = 1 + 2 = 3 For x = 2, y = (2)2 + 2 = 4 + 2 = 6 For x = 3, y = (3)2 + 2 = 9 + 2 = 11 Now, complete the above table with the calculated values of y x −2 −1 0 1 2 3 y 6 3 2 3 6 11

Figure 2.24: Graph of x² + 2 and x + 2 Now, plot the points and draw the quadratic graph as shown below.

c.

i. Since the graph of y = x² + 2 does not touch the x-axis, it has no solution (That is, it has complex roots).

ii. When x = 0, y = 2 and when x = 1, y = 3.

Finding the range of values of x for which y is increasing or decreasing

Figure 2.25: Maximum graph From the graph:

• The range of values of x for which y increases as x increases is x < 1.

• The range of values of x for which y decreases as x increases is x > 1.

• The range of values of x for which y is positive (greater than zero) is A < x < B.

• The range of values of x for which y is negative (less than zero) is or x < −0.8 or x > 2.8.

Figure 2.26: Minimum graph From the graph:

• The range of values of x for which y increases as x increases is x > 1.

• The range of values of x for which y decreases as x increases is x < 1.

• The range of values of x for which y is positive (greater than zero) is x < A or x > B.

• The range of values of x for which y is negative (less than zero) is A< x < B.

Example 2.27:

a. Copy and complete the following table for the relation y = x² + 2x − 1 for the interval −3 ≤ x ≤ 2 x −3 −2 −1 0 1 2 y

b. Using a scale of 1 cm to 1 unit on the x-axis and 1cm to 1 unit on y-axis, draw the graph of the relation.

c. From the graph, find:

i. The range of values of x for which y decreases as x increases.

ii. The range of values of x for which y increases as x increases.

iii. The range of values of x for which y is negative.

iv. The range of values of x for which y is positive.

Solution

a. Substitute the values into the equation y = x² + 2x − 1 to find the corresponding values of y.

For x = −3, y = (−3)2 + 2(−3) − 1 ⇒ y = 2 For x = −2, y = (−2)2 + 2(−2) – 1 ⇒ y = −1 For x = −1, y = (−1)2 + 2(−1) – 1 ⇒ y = −2 For x = 0, y = (0)2 + 2(0) − 1 ⇒ y = −1 For x = 1, y = (1)2 + 2(1) − 1 ⇒ y = 2 For x = 2, y = (2)2 + 2(2) − 1 ⇒ y = 7 The completed table is shown below.

x −3 −2 −1 0 1 2 y 2 −1 −2 −1 2 7

b. The graph is shown below.

Figure 2.27: Graph of x² + 2x − 1 c.

i. −3.8 ≤ x < −1

ii. −1< x ≤ 1.8

iii. −2.4 < x < 0.4

iv. x < −2.4 or x > 0.4

Example 2.28:

c.

i. x < −1

ii. x > −1

iii. A < x < B

iv. x < A or x > B Real-Life applications of simultaneous equations

Example 2.29:

A ball is thrown upward and its height after t seconds is given by: h = t² + 4t + 2 A drone is hovering at a constant height (h) of 5m.

Find the time when the ball is at the same height as the drone by drawing both graphs.

Complete the following table of values to help you solve this:

t 0 1 2 3 4 5 h

Solution

Substitute the values into the equation h = t² + 4t + 2 to find the corresponding values of h.

For t = 0, h = (0)2 + 4(0) + 2 ⇒ y = 2 For t = 1, h = (1)2 + 4(1) + 2 ⇒ y = 7 For t = 2, h = (2)2 + 4(2) + 2 ⇒ y = 14 For t = 3, h = (3)2 + 4(3) + 2 ⇒ y = 23 For t = 4, h = (4)2 + 4(4) + 2 ⇒ y = 34 For t = 5, h = (5)2 + 4(5) + 2 ⇒ y = 47 The completed table is shown below:

t 0 1 2 3 4 5 h 2 7 14 23 34 47 The graph is shown below.

Figure 2.28: Graph of h = t² + 4t + 2 and f(x) = 5 Therefore, the drone and the ball are at the same height when t = 0.65 seconds. As time cannot be negative we ignore the negative solution.

Activity 2.4 Drawing Graphs

1. Copy and complete the following table for the relation, y = x² − 4x +3 for −1 ≤ x ≤ 5.

x −1 0 1 2 3 4 5 y 8 −1 3

2. Using a scale of 2cm to 1 unit on both x and y axes, draw the graph of the relation using the completed table.

3. From your graph:

i. Write down the equation of the line of symmetry.

ii. State the range of values of x for which the curve is increasing.

4. State the range of values of x for which y > 0.

Activity 2.5 Real-Life Simultaneous Equation

Two taxi companies charge according to the equations:

• Company A: y = 2x + 4

• Company B: y = x² − 2x + 8 where y is the cost in cedis and x is the number of kilometres travelled.

1. Draw both graphs for 0 ≤ x ≤ 6 on the same set of axes using a table of values.

2. From your graph:

a. Find the number of kilometres at which both companies charge the same amount.

b. For what range of distances is Company B cheaper?

c. Discuss the results with your classmates.

1. Draw the graph of the quadratic function y = x² − 4x + 3, for −1 ≤ x ≤ 5

2. From the graph of y = − x² + 6 x − 5, determine its maximum value and the x-coordinates where it occurs.

Figure 2.29: Graph of y = −x² + 6x − 5

3. Draw the graph of y = x² − 2x − 8 and use it to determine the roots of the equation. Explain how the graph shows these roots.

4. Without solving algebraically, explain how you can determine whether the equation, 2x² − 3x + 4 = 0 has complex roots by using its graph.

5. A ball is thrown upward and its height (in metres) after t seconds is given by h(t) = −5t²+ 20t + 2. Draw the graph of h(t) for and use it to determine the maximum height of the ball and the time at which this occurs.

6. Draw and compare the graphs of y = x² − 4x + 3 and y = −x² + 4x − 3.

Describe in detail how their shapes, maximum/minimum points, and roots differ, and explain why.

7. Solve the quadratic equation: x² − 5x + 6 = 0

8. Solve the quadratic equation 2x² + 3x – 2 = 0 using the quadratic formula.

9. Given the graph of the quadratic function y = −x² + 4x + 1, determine the coordinates of the maximum point.

10. Sketch the graph of y = x² − 6x + 8. Identify and label its vertex and intercepts.

11. Write the equation of the axis of symmetry for y = 2x² − 8x + 3.

12. A ball is thrown into the air, and its height is given by the equation h(t) = −5t² + 20t + 1

a. What is the maximum height?

b. After how many seconds does it hit the ground?

13. Using the graph of y = −x² + 4x − 3, determine the values of x for when y is positive.

14. For the function y = x² − 6x + 5, use the graph to determine the x value interval(s) where:

a. y is decreasing

b. y is increasing

15. A company’s profit P(x) in thousands of dollars is modelled by the equation P(x) = −2x² + 12x − 5, where x is the number of items produced in hundreds.

a. What number of items will maximize profit?

b. What is the maximum profit?

c. For what values of x is the profit positive?

16. A rectangular garden is to be fenced on three sides with a fixed length of fencing: 60 metres, with one side left open along a wall.

Let the width of the garden be x meters.

a. Write an expression for the length of the garden in terms of x.

b. Write the area of the garden as a function of x.

c. Determine the value of x that will maximize the area and find the maximum area.

d. Interpret your solution in the context of the problem.

17. A projectile is launched from a platform, and its height (in metres) after t seconds is modelled by the equation: h(t) = −4.9t² + 14t + 3

a. Determine the time when the projectile reaches its maximum height and find the maximum height.

b. Find the total time of flight before it hits the ground.

c. A sensor is placed at a height of 10 metres.

d. Explain how you can verify your answers graphically and describe the shape and key features of the graph.

18. Two companies offer different payment plans, where x represents the hours worked:

• Company A charges a flat rate: y = 25x + 100

• Company B charges based on a quadratic model: y = 5x² + 10x

a. Graph both functions on the same set of axes.

b. Determine the number of hours for which both companies charge the same amount.

c. For what range of hours is Company B cheaper than Company A?

d. Evaluate the practicality of each plan depending on the number of hours and explain which option is better in different scenarios.