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Gradient and Intercept 06 October 2015 Lesson Objective: To Plot the graphs of simple linear functions, and also find the equation of a straight line -

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Presentation on theme: "Gradient and Intercept 06 October 2015 Lesson Objective: To Plot the graphs of simple linear functions, and also find the equation of a straight line -"— Presentation transcript:

1 Gradient and Intercept 06 October 2015 Lesson Objective: To Plot the graphs of simple linear functions, and also find the equation of a straight line - L5 Lesson Outcomes Must: plot graphs given equations of the line – D Should: find equation of line –D/C Could: Find the equation given the graph– C

2 Gradients of straight-line graphs The gradient of a line is a measure of how steep the line is. y x a horizontal line Zero gradient y x a downwards slope Negative gradient y x an upwards slope Positive gradient The gradient of a line can be positive, negative or zero if, moving from left to right, we have If a line is vertical its gradient cannot by specified.

3 The equation of a straight line y = mx + c m = gradient of the line. It tells us how steep the line is or the SLOPE of the line c = intercept on the y axis tells us where the line crosses the y-axis. 3 5 1 4 -5-4-3-2 4 3 2 -2 -4 -3 -6 -5 0 21 6 = Gradient is 1 1 = 1 Intercept is 2 y = 1x + 2 = Gradient is 3 1 = 3 Intercept is 3 y = 3x + 3 = Gradient is 0.5 1 = 0.5 Intercept is 1 y = 0.5x + 1

4 The gradient and the y-intercept equationgradienty-intercept y = 3x + 4 y = – 5 y = 2 – 3x 1 –2–2 3(0, 4) (0, –5) –3 (0, 2) y = x y = –2x – 7 x 2 1 2 (0, 0) (0, –7)

5 Finding the gradient from two given points If we are given any two points ( x 1, y 1 ) and ( x 2, y 2 ) on a line we can calculate the gradient of the line as follows, the gradient = change in y change in x the gradient = y 2 – y 1 x 2 – x 1 x y ( x 1, y 1 ) ( x 2, y 2 ) y 2 – y 1 Draw a right-angled triangle between the two points on the line as follows,

6 Draw a Graph Plot the points and write the equation Level 7/8 Pack 1

7 Sketching Graphs Y = x + 3 Y = -x + 2 The sign in front of the x tells you if the gradient is positive or negative. Positive gradient Negative gradient This tells you where the line cuts the vertical line 3 cuts the vertical line at 2 2

8 1. y = x + 5 2. y = x + 1 3. y = x + 3 4. y = x - 6 5. y = x - 5 6. y = x - 1 7. y = -x + 7 8. y = -x + 3 9. y = -x - 4 10. y = 2x + 5 Starter: sketch the graph of each of the equation. 5 1. 1 2. 3 3. -6 4. -5 5. 6. 3 8. -4 9. 7 7. 5 10.

9 Linear graphs with positive gradients

10 Investigating straight-line graphs

11 Rearranging equations into the form y = mx + c Sometimes the equation of a straight line graph is not given in the form y = mx + c. The equation of a straight line is 2 y + x = 4. Find the gradient and the y -intercept of the line. We can rearrange the equation by transforming both sides in the same way 2 y + x = 4 2 y = – x + 4 y = – x + 4 2 y = – x + 2 1 2

12 Rearranging equations into the form y = mx + c Sometimes the equation of a straight line graph is not given in the form y = mx + c. The equation of a straight line is 2 y + x = 4. Find the gradient and the y -intercept of the line. Once the equation is in the form y = mx + c we can determine the value of the gradient and the y -intercept. So the gradient of the line is 1 2 – and the y -intercept is 2. y = – x + 2 1 2

13 What is the equation? Look at this diagram: What is the equation of the line passing through the points a) A and E b) A and F c) B and E d) C and D e) E and G f) A and C? x = 2 y = 10 – x y = x – 2 y = 2 y = 2 – x y = x + 6 x y

14 Substituting values into equations What is the value of m ? To solve this problem we can substitute x = 3 and y = 11 into the equation y = mx + 5. This gives us: 11 = 3 m + 5 6 = 3 m Subtracting 5: 2 = m Dividing by 3: m = 2 The equation of the line is therefore y = 2 x + 5. A line with the equation y = mx + 5 passes through the point (3, 11).

15 Pairs

16 Matching statements

17 Exploring gradients


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