Tuesday, February 19, 2019

yl: mathematics

Algebra

Clement Radcliffe CONTRIBUTOR W E CONTINUED, last week, the review of Algebra. Much time has been spent on this topic, and I do recommend that you strive for mastery in all areas. Again, I am urging you to proceed to study with systematic and ongoing practice. Let us now continue the review of graphs. GRAPHS We will now complete the review of graphs with an illustration of the concepts which we reviewed. 1. Given the graph of the function f(x) = 2x 2 – 9x – 5, solve:

Answer : m = 1.

PRACTICE EXAMPLE Find the gradient of the line the following points. a. A ( 2 , 4) B ( 4 , 8) b. X ( 3 , – 4) Y (– 5 , 0) Please note that the correct answer for b. is: m = –1/2 INTERCEPT This is the y: coordinate of the point, where the line cuts the y axis, that is the point (o , y). This y value is denoted as c. The following is a plot of the points A and B on the Cartesian plane which will illustrate the concepts. MIDPOINT Given the points A(x1, y1) & B( x 2 , y 2 ), then the midpoint is equal distance from A and B. This point is denoted by M and from the diagram, the coordinates of the midpoint are: EXAMPLE Find the coordinates of M, mid-point of AB. Using the coordinates A (3 , 1) , B (6 , 4). Given above, M = 6 + 3 , 4 + 1 2 2 ( ) ( ) Answer: 9 , 5 2 2 In review: Given the points A(x 1 , y 1 ), then finding the gradient and midpoint involves substi- tuting into the appropriate formula. This is illustrated as follows: EXAMPLE Given the points A(6 , –3) and B( – 4 , 1), find: (i) The gradient of AB (ii) The midpoint of AB ) and B(x 2 , y 2 M = x 2 + x ( ) 1 , y 2 + y 1 2 2

x – 2 – 1 0 1 2 3 4 y 31 11 – 1 – 5 – 1 11 31

n The minimum value is – 5 n The minimum value is at x = 1 n The equation of the axis of symmetry is: x = 1. We will now begin to review coordinate geometry by considering straight lines on the Cartesian plane with respect to:

n Gradient n Intercept n Midpoint n Length n Equation

ii. 2x 2 – 9x – 5 = 0 iii.2x 2 – 10x – 7 = 0

Please explain your method.

Again, let me remind you of the importance of the theory of graphs, as it is very important to this topic. The Cartesian plane consists of the perpendicular x and y axes. REMINDERS n The axes must be properly labelled. n Appropriate scales should be accurately used. n If the scales with respect to the axes are given, then they must be used as given. n The axes usually cross at the point (0 , 0). n The coordinates of a point are always expressed in the form: (x , y). n Points are usually named with capital letters, for example, P(x , y). n Three points are required to draw a straight line. A ruler MUST ALWAYS be used to join the points. GRADIENT The gradient of a line is a measure of its slope. The value is denoted by m and given two points on the slope; it is defined as: m = Increase in the y coordinates Increase in the x coordinates Given that the two points are represented by A (x1 , y1), and B (x2, y2), then the formula is: m = y 2 – y1 x 2 – x1 EXAMPLE Find the gradient m of the line joining the points A (3 , 1) , B (6 , 4). Since m = y 2 – y1, substituting x 2 – x1 m = 4 – 1 = 3 = 1. 6 – 3 3 Please be sure to substitute in the correct order.

SOLUTION

i. y = x + 2 x 0 1 2 y 2 3 4

y = 2x 2 – 9x – 5 x – 1 0 1 2 3 4 5 6 y 6 – 5 – 12 –15 –14 –9 0 13

ii. If 2x 2 – 9x – 5 = 0, then x = – 1 Or 5 (Points where the curve cuts the x axis). iii. If 2x 2 – 10x – 7 = 0, then you reorganise this equation so that the expression 2x 2 – 9x – 5 is on one side, that is: 2x 2 – 10x – 7 = 2x 2 – 9x – x –5 – 2 = 0 (– 10x = – 9x – x and – 7 = –5 – 2) 2x 2 – 9x – x –5 – 2 = 2x 2 – 9x – 5 – x – 2 = 0 2x 2 – 9x – 5 = x + 2 The solution of the equation 2x 2 – 10x – 7 = 0 is the same as that of the equation 2x 2 – 9x – 5 = x + 2 By plotting the line y = x + 2, and read off the coordinates of the points of intersection with the curve, then: x = – 0.6 Or 5.7

SOLUTION (i) Gradient of AB = m = y 2 Substituting the coordinates x 2 m = 1 – –3 = 1 + 3 = –4. –4 – 6 – 10 10 m = –2/5

– y

2. Given that h(x) = 4x 2 – 8x – 1

1

– x

1

By plotting the function h(x), find: n Its minimum value. n The value of x for which h(x) is a minimum. n The equation of the axis of symmetry.

+ x ( ) CONTINUED ON PAGE 24 1 , y 2 + y 1 2 2

(ii) The midpoint of AB = M = x 2

SOLUTION

y = 4x 2 – 8x – 1

20

YOUTHLINK MAGAZINE | www.youthlinkjamaica.com | FEBRUARY 19 - FEBRUARY 25, 2019