2The Video Shop.You join a video shop for a membership fee of £3 and then charges £2 for each video you hire:Complete the table below for the cost of hiring different numbers of videos.No of videosCost of Videos(£)3579111315Now draw a graph of the table above.
31432567891011121314No of VideosCostGraph of videos hired against cost.
41432567891011121314No of VideosCostNow consider the structure of the graph.The graph cuts the y axis at (0,3) because it cost £3 to join the video shop before you hired any videos.For every square that you move to the right you go two squares up because the cost of each video is £2.
5Finding A Formula.Look at the table of values for the video hire once again:VC3579111315Find a formula for the cost of videos (C) given the number of videos (V) :C =2V + 3This was the number of squares you went up for each one you went along.This is called the gradient of the line.This was the place were the graph cut the y axis. This is called the y axis intercept.From the graph we saw that :
6Now repeat the question again for a video shop charging £5 to join and £3 for each video hired.Start by completing the table below.No of videosCost of Videos(£)5Answer the questions below:Where does the graph cut the y axis ?( 0 , 5 )What is the gradient of the line:Gradient = 3The full graph is shown on the next slide:
7Gradient of 3.Y axis intercept (0,5)1432567891011121314No of VideosCostC = 3V + 5Is the equation of the line.
8More About The Gradient. The gradient (m) of a straight line is defined to be:Change in vertical height.Change in horizontal distance.We are going to use this definition to calculate the gradient of various straight lines:
9Find the gradients of the straight lines below: (1)(2)47m =4374m =34(3)44m =44m = 1
11Negative Gradient Consider the straight lines shown below: (d) (e) (a) Can you split the lines into two groups based on their gradients ?Positive gradientLines (a) (c) and (d) slope upwards from left to right.Negative gradientLines (b) and (e) slope downwards from left to right.
12Calculate the gradients of the lines below: (1)(2)- 4- 856
13The Equation Of A Straight Line. To find the equation of any straight line we require to know two things:(a) The gradient of the line.m = gradient.(b) The y axis intercept of the line.c = y axis intercept.The equation of a straight line is :y = m x + cExamples.Give the gradient and the y axis intercept for each of the following lines.(1) y = 6x + 5(2) y = 4x + 2(3) y = x - 3m = 6c = 5m = 4c = 2m = 1c = - 3
14Finding The Equation. Find the equation of the straight lines below: x y(1)xy(2)What is the gradient ?m = 1What is the y axis intercept?c = 2c = 1Now use y = m x + cy = x + 2
17The Gradient Formula. The Gradient Formula. It shows a straight line passing through the points (x1,y1) and (x2,y2).Look at the diagram below:x1x2y1y2We must calculate the gradient of the line using the triangle shown:y2-y1Change in vertical height:y2 – y1x2-x1Change in horizontal distance:x2-x1Gradient of line:
18Calculate the gradient of the line through the points below: (2) C(-4,8) and D(6,-10)(1) A(4,6) and B( 10,12)Solution:Solution:Write down the gradient formula:Gradient of line:Substitute in your values:Calculate and simplify:
19Straight Line From Two Points. Find the equation of the straight line passing through (4,6) and (8,12)Solution:Now substitute one of the points into y = m x + c to find c.Find the gradient of the line:Sub’ (4,6) into y = 3x +c :6 = 3 x 4 +cc + 12 = 6c = - 6Substitute gradient into y = m x + c.Now write down the equation of the straight line:y = 3 x + cy = 3x - 6
20Find the equation of a straight line passing through C(6,-7) and D(-12,9) Solution.Now substitute one of the points into y = m x + c to find c.Calculate the gradient:Sub’ (6,-7) into equation:Substitute gradient into y = m x + c.c = -23Equation of the straight line: