2.1 Graphs of Functions (Part 3)

Posted on May 27, 2020 by Myhometuition

(B) Types of graphs of functions.

(c) Cubic function: y = ax³ + bx + c

Highest power of the variable x is 3.

(i) y = ax³ + bx + c, a> 0

(ii) y = ax³ + bx + c, a< 0

(d) Reciprocal function:

y = \frac{a}{x}

Highest power of the variable x is –1.

2.5 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 6:
(a) Complete the table in the answer space for the equation

y = \frac{36}{x}

by writing down the values of y when x = 3 and x = 8.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 1 cm to 1 unit on the x-axis and 1 cm to 1 unit on the y-axis, draw the graph of

y = \frac{36}{x}

for 2 ≤ x ≤ 14.

(c) From your graph, find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 4.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation

\frac{36}{x} + x - 14 = 0

for 2 ≤ x ≤ 14.

Answer:

x	2	2.4	3	4	6	8	10	12	14
y	18	15		9	6		3.6	3	2.6

Solution:
(a)

\begin{array}{l} y = \frac{36}{x} \\ when x = 3 \\ y = \frac{36}{3} = 12 \\ when x = 8 \\ y = \frac{36}{8} = 4.5 \end{array}

(b)

(c)
(i) From the graph, when x = 2.6, y = 13.6
(ii) From the graph, when y = 4, x = 9

(d)

\begin{array}{l} y = \frac{36}{x} ........... (1) \\ 0 = \frac{36}{x} + x - 14 ........... (2) \\ (1) - (2) : \\ y = - x + 14 \end{array}

The suitable straight line is y = –x + 14.

x	2	12
y = –x + 14	12	2

From the graph, x = 3.4, 10.6.

2.5 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 5:

(a) The following table shows the corresponding values of x and y for the equation

y = \frac{24}{x}

x	–4	–3	–2	–1	1	1.5	2	3	4
y	–6	k	–12	–24	24	n	12	8	6

Calculate the value of k and n.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.

By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of

y = \frac{24}{x}

for –4 ≤ x ≤ 4.

(c) From your graph, find

(i) The value of y when x = 3.5,

(ii) The value of x when y = –17.

(d) Draw a suitable straight line on your graph to find the value of x which satisfy the equation 2x² + 5x = 24 for –4 ≤ x ≤ 4.

Solution:

(a)

\begin{array}{l} y = \frac{24}{x} \\ when x = - 3, \\ k = \frac{24}{- 3} = - 8 \\ when x = 1.5, \\ n = \frac{24}{1.5} = 16 \end{array}

(b)

(c)

(i) From the graph, when x = 3.5, y = 7

(ii) From the graph, when y = –17, x = –1.4

(d)

\begin{array}{l} y = \frac{24}{x} -------------- (1) \\ 2 x^{2} + 5 x = 24 ------ (2) \\ (2) \div x, 2 x + 5 = \frac{24}{x} -------- (3) \\ (1) - (3), y - (2 x + 5) = 0 \\ y = 2 x + 5 \end{array}

The suitable straight line is y = 2x + 5.

Determine the x-coordinate of the point of intersection of the curve and the straight line y = 2x + 5.

x	0	3
y = 2x + 5	5	11

From the graph, x = 2.5

2.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

2.4 SPM Practice (Short Questions)

Question 1:

Which of the following graphs represent y = –x²– x + 2?

Solution:

1. The coefficient of x²< 0, its shape is ∩.

2. Find the x-intercepts by substituting y = 0 into the function.

y = –x² –x + 2

when y = 0

0 = (–x + 1) (x + 2)

x = 1, –2

The answer is A.

Question 2:

Which of the following graphs represent 3y = 18 + 2x?

Solution:

1. The highest power of x = 1, it is a linear graph, i.e. a straight line.

2. The coefficient of x > 0, the straight line is /.

3y = 18 + 2x

On the y-axis, x = 0.

3y = 18

y = 6

On the x-axis, y = 0.

0 = 18 + 2x

2x = –18

x = –9

The answer is B.

2.2 Solving Equations Graphically

Posted on April 24, 2020 by user

2.2 Solving equations graphically

The solution of the equation f (x) = g (x) can be solved by graphical method.

Step 1: Draw the graphs of y = f (x) and y = g (x) on the same axes.

Step 2: The points of intersection of the graphs are the solutions of the equation f (x) = g(x). Read the values of x from the graph.

Solution of an Equation by Graphical Method

Example 1:

(a) The following table shows the corresponding values of x and y for the equation y = 2x² – x – 3.

x	–2	–1	–0.5	1	2	3	4	4.5	5
y	7	m	– 2	–2	3	12	n	33	42

Calculate the value of m and n.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.

By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x² – x – 3 for –2 ≤ x ≤ 5.

(c) From your graph, find

(i) The value of y when x = 3.9,

(ii) The value of x when y = 31.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation 2x²– 3x = 10 for –2 ≤ x ≤ 5.

Solution:

(a)

y = 2x² – x – 3

when x = –1,

m = 2 (–1)² – (–1) – 3 = 0

when x = 4,

n = 2 (4)² – (4) – 3 = 25

(b)

(c)

(i) From the graph, when x = 3.9, y = 23.5

(ii) From the graph, when y = 31, x = 4.4

(d)

y = 2x² – x – 3 ----- (1)

2x²– 3x = 10 ----- (2)

y = 2x² – x – 3 ----- (1)

0 = 2x²– 3x – 10 ------ (2) ← (Rearrange (2))

(1) – (2) : y = 2x + 7

The suitable straight line is y = 2x + 7.

Determine the x-coordinates of the two points of intersection of the curve

y = 2x² – x – 3 and the straight line y = 2x+ 7.

x	0	4
y = 2x + 7	7	15

From the graph, x = –1.6, 3.1

2.1 Graphs of Functions (Part 2)

Posted on April 24, 2020 by user

(B) Types of graphs of functions.

(a) Linear function: y = mx + c

Highest power of the variable x is 1.

(i) y = mx + c, m > 0

(ii) y = mx + c, m < 0

(b) Quadratic function: y = ax² + bx + c

Highest power of the variable x is 2.

(i) y= ax² + bx + c, a > 0

(ii) y= ax² + bx + c, a < 0

2.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

2.4.2 SPM Practice (Short Questions)

Question 3:

Which of the following graphs represent y = 2x³– 16?

Solution:

y = 2x³ – 16

On the y-axis, x = 0.

y = 2(0)³ – 16

y = –16

The answer is C.

Question 4:

Which of the following graphs represent

y = - \frac{3}{x} ?

Solution:

y = - \frac{3}{x} or y = - 3 x^{- 1}

Highest power of the variable x is –1.

It is a reciprocal function

y = \frac{a}{x}, in this case a = - 3.

The answer is D.

2.1 Graphs of Functions (Part 1)

Posted on April 24, 2020 by user

2.1 Graphs of Functions

(A) Drawing graphs of functions

1. To draw the graph of a given function, the following steps are taken.

Step 1: Construct a table of values for the given function.

Step 2: Select a suitable scale for the x-axis and y-axis, if it is not given.

Step 3: Plot the points and complete the graph by joining the points.

Step 4: Label the graph.

2.3 Region Representing inequalities in Two Variables (Part 1)

Posted on April 24, 2020 by user

2.3.1 Position of a point relative to the graph of y = ax+ b

1. When y = ax + b, the point is on the line.

2. When y < ax + b, the point is below the line.

3. When y > ax + b, the point is above the line.

2.3.2 Identifying the region satisfying the respective inequalities.

1. A dashed line ‘----’ is used when points on the line are not included in the region representing the inequality, such as y > ax + b or y < a or x > a.

2. A solid line ‘___’ is used when points on the line are not included in the region representing the inequality, such as y ≥ ax + b or y ≤ a or x ≥ a.

3. The diagrams below show the regions that satisfy the respective inequalities.

2.3.3 Region Representing inequalities in Two Variables (Sample Questions)

Posted on April 24, 2020 by user

Example 1:

On the graph in the answer space, shade the region which satisfies the three inequalities y ≥ –x + 10, y > x and y < 9.

Answer:

Solution:

1. The region that satisfies the inequality y ≥ –x + 10 is the region on and above the line y = –x + 10.

2. The region that satisfies the inequality y > x is the region above the line y = x. Since inequality sign “ >” is used, the points on the line y = x is not included. Thus, a dashed line needs to be drawn for y = x.

3. The region that satisfies the inequality y < 9 is the region below the line y = 9 drawn as a dashed line.

Example 2:

On the graph in the answer space, shade the region which satisfies the three inequalities y ≥ –2x + 10, y ≤ 10 and x < 5.

Answer:

Solution:

1. The region which satisfies the inequality of the form y ≥ ax + c is the region that lies on and above the line y = ax+ c.

2. In this question, y intercept, c = 10, x intercept is 5.

3. The region that satisfies the inequality y ≥ –2x + 10 is the region on and above the line y = –2x + 10.

4. The region that satisfies the inequality y ≤ 10 is the region on and below the line y = 10.

5. The region that satisfies the inequality x < 5 is on the left of the line x = 5 drawn as a dashed line.