GRAPHING LINEAR EQUATIONS USING A TABLE OF VALUES

If we are given an equation we can plot a graph using the following procedure :

Draw the table with 5 different values that fit the given equation.
By applying random values of x, we can apply the values of y.
Plot the points on a number grid.
Join the points with a straight line.
Place arrows at the ends to indicate that the line extends in both directions.

Copy and complete the following tables for the equations provided. Plot each set of ordered pairs on separate axes and draw straight lines through the points.

Example 1 :

Solution :

x = -2

y = -2

x = -1

y = -1

x = 0

y = 0

x = 1

y = 1

x = 2

y = 2

So, the points are (-2, -2) (-1, -1) (0, 0) (1, 1) and (2, 2).

By plotting the points in the graph paper, we will get the graph given above.

Example 2 :

Solution :

If x = -4

y = -4+3

y = 1

If x = -2

y = -2+3

y = 1

If x = 0

y = 0+3

y = 3

If x = 2

y = 2+3

y = 5

If x = 4

y = 4+3

y = 7

So, the points are (-4, -1) (-2, 1) (0, 3) (2, 5) and (4, 7). By plotting the points, we get

Example 3 :

Solution :

If x = -2

y = 4-(-2)

y = 6

If x = -1

y = 4-(-1)

y = 5

If x = 0

y = 4-0

y = 4

If x = 1

y = 4-1

y = 3

If x = 2

y = 4-2

y = 2

So, the points are (-2, 6) (-1, 5) (0, 4) (1, 3) and (2, 2). By plotting the points, we get

Example 4 :

Solution :

If x = -2

y = 2(-2)-2

y = -4-2

y = -6

If x = -1

y = 2(-1)-2

y = -2-2

y = -4

If x = 0

y = 2(0)-2

y = 0-2

y = -2

If x = 1

y = 2(1)-2

y = 2-2

y = 0

If x = 2

y = 2(2)-2

y = 4-2

y = 2

So, the points are (-2, -6) (-1, -4) (0, -2) (1, 0) and (2, 2). By plotting the points, we get

Example 5 :

Solution :

If x = -2

y = 2(-2)-2

y = -4-2

y = -6

If x = -1

y = 2(-1)-2

y = -2-2

y = -4

If x = 0

y = 2(0)-2

y = 0-2

y = -2

If x = 1

y = 2(1)-2

y = 2-2

y = 0

If x = 2

y = 2(2)-2

y = 4-2

y = 2

So, the points are (-2, 7) (-1, 5) (0, 3) (1, 1) and (2, -1). By plotting the points, we get

Example 6 :

Solution :

If x = -2

y = 4(-2)-5

y = -8-5

y = -13

If x = -1

y = 4(-1)-5

y = -4-5

y = -9

If x = 0

y = 4(0)-5

y = 0-5

y = -5

If x = 1

y = 4(1)-5

y = 4-5

y = -1

If x = 2

y = 4(2)-5

y = 8-5

y = 3

So, the points are (-2, -13) (-1, -9) (0, -5) (1, -1) and (2, 3). By plotting the points in the graph, we get

Example 7 :

Solution :

If x = -2

y = 7-3(-2)

y = 7+6

y = 13

If x = -1

y = 7-3(-1)

y = 7+3

y = 10

If x = 0

y = 7-3(0)

y = 7 - 0

y = 7

If x = 1

y = 7-3(1)

y = 7-3

y = 4

If x = 2

y = 7-3(2)

y = 7-6

y = 1

So, the points are (-2, 13) (-1, 10) (0, 7) (1, 4) and (2, 1).. By plotting these point in the graph, we get

Example 8 :

(a) Draw y = 3x − 4

(b) Draw x + y = 2

The graph y = 3x − 4 crosses the y-axis at the point A The graph x + y = 2 crosses the x-axis at the point B O is the origin.

(d) Write down the coordinates of the point B

(e) Find the area of triangle OAB

Solution :

y = 3x − 4

When x = -2

y = 3(-2) - 4

= -6 - 4

y = -10

When x = -1

y = 3(-1) - 4

= -3 - 4

y = -7

When x = 0

y = 3(0) - 4

= 0 - 4

y = -4

When x = 1

y = 3(1) - 4

= 3 - 4

y = -1

When x = 2

y = 3(2) - 4

= 6 - 4

y = 2

When x = 3

y = 3(3) - 4

= 9 - 4

y = 5

So, the points are (-2, -10)(-1, -7)(0, -4) (1, -1) (2, 2) and (3, 5)

x + y = 2

y = 2 - x

When x = -2

y = 2 - x

= 2 + 2

y = 4

When x = -1

y = 2 - x

= 2 + 1

y = 3

When x = 0

y = 2 - x

= 2 - 0

y = 2

When x = 1

y = 2 - x

= 2 - 1

y = 1

When x = 2

y = 2 - x

= 2 - 2

y = 0

When x = 3

y = 2 - x

= 2 - 3

y = -1

So, the points are (-2, 4) (-1, 3) (0, 2) (1, 1) (2, 0) and (3, -1)

Area of triangle OAB = (1/2) x base x height

= (1/2) x 4 x 2

= 4 square units.

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GRAPHING LINEAR EQUATIONS USING A TABLE OF VALUES

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