Unit 4:Mathematics Aims Introduce linear equations. Objectives Use algebra formula to plot straight line graphs.

y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Example 1 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4 X = 3

Example 2 y 4 3 2 1 -4 -3 -2 -1 y = -1 x -4 -3 -2 -1 0 1 2 3 4

Question 1 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Question 2 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Question 3 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Question 4 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Question 5 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Question 6 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Question 7 y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

Straight-line graphs x -3 -2 -1 1 2 3 Y=x+3 4 5 x = -3 y = -3 + 3 = 0 Complete the table of values for y = x + 3, and then draw the graph from -3 to 3 for x x = -3 y = -3 + 3 = 0 x = -2, y = -2 + 3 = 1 x = 0 y = 0 + 3 = 3 x = 3 y = 3 + 3 = 6 x -3 -2 -1 1 2 3 Y=x+3 4 5

y 4 3 2 1 -4 -3 -2 -1 x -4 -3 -2 -1 0 1 2 3 4

COORDINATE PLANE Y-axis X-axis Parts of a plane X-axis Y-axis Origin Quadrants I-IV Y-axis QUAD II QUAD I Origin ( 0 , 0 ) X-axis QUAD III QUAD IV

PLOTTING POINTS Remember when plotting points you always start at the origin. Next you go left or right. Then you go up or down. B C Plot these 4 points A (3, -4), B (5, 6), C (-4, 5) and D (-7, -5) A D

L SLOPE Run is 6 because we went to the right Rise is 10 because we went up Rise is -10 because we went down Run is -6 because we went to the left

All straight lines can be written in the form y = mx + c You need to be able to write down the equation of a straight line by working out the values for m and c.

c is the constant value – this part of the function does not change. y = mx + c m is the gradient of the line This type of equation was made popular by the French Mathematician Rene Descartes. “m” could stand for “Monter” – the French word meaning “to climb”.

y x Finding m and c y = mx +c 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 –7 –6 –5 –4 –3 –2 –1 -1 -2 -3 -4 -5 -6 It is very easy to find the value of c – this is the point at which the line crosses the y-axis. Each time the lines moves 1 place to the right, it climbs up by 2 places. So m = 2 Finding m is also easy in this case. The gradient means the rate at which the line is climbing. y = mx +c y = 2x +3

As the line travels across 1 position, it is not clear how far up it has moved. y x 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 –7 –6 –5 –4 –3 –2 –1 -1 -2 -3 -4 -5 -6 y = mx +c But… any right angled triangle will give us the gradient! Let’s draw a larger one. In general, to find the gradient of a straight line, we divide the… vertical change by the… horizontal change 2 y = mx +c y = ½x - 2 4 The gradient, m = 2/4 = ½

Equation of a straight-line graph The y-intercept of a graph is the y-coordinate of the point where the graph cuts the y-axis. It is usually denoted by c.

Equation of a straight-line graph. Exercise. Find the y-intercept of the following: 2)

Equation of a straight-line graph. The gradient, steepness, of a line and its y-intercept define a line uniquely. In fact any straight line has an equation of the form: y = mx + c, where m is the gradient and c is the y-intercept. Example. Find the equation of the line with gradient 3 and y-intercept 4. Here m = 3 and c = 4. We substitute the values of m and c into the equation to obtain: y = 3x + 4.

Equation of a straight-line graph. Exercise. Find the equations of the straight lines with the following gradients and y-intercepts: 1. Gradient of 4 and y-intercept of 2. 2. Gradient of –6 and y-intercept of 8 3. Gradient of ½ and y-intercept of –7. (Answers: y = 4x + 2; y = -6x+8; y = ½ x - 7.)

Equation of a straight-line graph We frequently need to find the equation of a line directly from its graph. We should: Find the y-intercept, c. Find the gradient, m, of the line. Substitute these values into y = mx + c. First, we find the y-intercept: c = 3. Then we find the gradient: m =

Equation of a straight-line graph. Example. Find the equation of the line that goes through the points (2, 4) and (6, 12). Choosing the point (6, 12) and substituting gives: y = m χ x+ c 12 = 2 x 6 + c. Rearranging gives: c = 0. Hence y = 2x + 0 y = 2x.

FORMULA FOR FINDING SLOPE Exercise. Find the equations of the lines that pass through the following points: 1. (0, 7) and (3, 13). 2. (2, 1) and (5, -1). 3. (4, 8) and (8, 0). 4. (-4, 2) and (12, 10). 1)Y=2x+7 2)y=-2 x + 7 3 3 3) y=-2x+16 4) y=1/2 x + 4