Reviewing algebra
WE COMPLETED, last week, the review of algebra. Much time was spent on this and I do recommend 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 9x
2 - 5, solve: ii) 2x2 - 9x - 5 = 0 iii) 2x2 - 10x - 7 = 0
SOLUTION GRAPH 1
From the graph:
(ii) If 2x2 - 9x - 5 = 0, then x = - 1 or 5 (Points where the curve cuts the x axis).
(iii) If 2x2 - 10x - 7 = 0, then you reorganise this equation so that the expression,
2x2 - 9x - 5 is on one side, that is:
2x2 - 10x - 7 = 2x2 - 9x - x -5 - 2 = 0
(- 10x = - 9x - x and - 7 = -5 - 2)
2x2 - 9x - x -5 - 2 = 2x2 - 9x - 5 - x - 2 = 0 2x2 - 9x - 5 = x + 2
The solution of the equation 2x2 - 10x - 7 = 0 is the same as that of the equation 2x2 - 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 2. Given that h(x) = 4x2 - 8x - 1 By plotting the function h(x), find :
Its minimum value.
The value of x for which h(x) is a minimum.
The equation of the axis of symmetry.
SOLUTION
y = 4x2 - 8x - 1
GRAPH 2
The minimum value is - 5
The minimum value is at x = 1
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:
Exploring the following aspects of a straight line: Gradient Intercept Midpoint Length Equation
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
The axes must be properly labelled. Appropriate scales should be accurately used.
If the scales, with respect to the axes are given, then they must be used as given.
The axes usually cross at the point (0 , 0). The coordinates of a point are always expressed in the form: (x , y).
Points are usually named with capital letters, for example, P(x , y).
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 = y2 - y1 / x2 - x1
EXAMPLE
Find the gradient m of the line joining the points A (1 , 2) , B (5 , 4).
Since m = y2 - y1 / x2 - x1, substituting m = 4 - 2/ 5-1 = 2/4 = 1/2.
Please be sure to substitute in the correct order.
Answer: 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.
GRAPH 3 MIDPOINT
Given the points A(x1, y1) & B( x2, y2), then the midpoint is equal distance from A & B. This point is denoted by M and from the diagram, the coordinates of the midpoint are:
M = x2 + x1 /2, y2 + y1/2
EXAMPLE
Find the coordinates of M, midpoint of AB. Using the coordinates of A and B given above,
M = 2 + 3/2 , 4 + 5/2 Answer: 5/2, 9/2
IN REVIEW
Given the points A(x1 , y1) and B(x2 , y2), then finding the gradient and midpoint involves substituting 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
SOLUTION
(i) Gradient of AB = m = y2 - y1 / x2 - x
1 Substituting the coordinates m = 1 - -3 /-4 -6 = 1 + 3/-10 = -4/10. m = -25
(ii) The midpoint of AB =
M = x2 + x1 /2, y2 + y1 /2 Substituting
M = 6 - 4 /2, -3 + 1 /2 = 2 /2 , - 2/2 M =(1 , -1 )
HOMEWORK
(1) Given the points A (2 , - 3) and B( 4 , - 5), find the values of:
(a) m( Gradient) (b) M( midpoint)
(2) The line K passes through the points A (6, 6) and B (0, - 2).
Find: (i) The midpoint of the AB.
(ii) The gradient of the line K.
(3) The line segment connects the points M (1, 8) and N (r, s). If the midpoint of MN is (4, 5), calculate the values of r and s.
(4) Given the points X ( 1 , 0) and Y (- 2 , a ), if the gradient is 2/3, find a.
Have a good week.
Clement Radcliffe is an independent contributor. Send questions and comments to kerry-ann.hepburn@gleanerjm.com