Question 1: Match up the equation with its graph Starter y = x2 y = -x2 + 2 y = -x2 - 2 y = x2 - 2 y = x2 + 2 y = -x2 a) e) 5 5 b) -5 c) Question 2: Write the gradients of each of these lines 5 -5 5 -5 -5 5 10 d) f) -5 5 -5 -5 5 -10 Question 1: Match up the equation with its graph Starter y = x2 y = -x2 + 2 y = -x2 - 2 y = x2 - 2 y = x2 + 2 y = -x2 a) Question 2: Write the gradients of each of these lines 5 1 4 b) 2 -5 5 -5 Gradient = -1 Gradient = 2 d) 5 4 e) f) 5 -5 -5 c) 5 1 10 1 -5 5 2 -5 5 8 -5 Gradient = 1/4 -10 Gradient = -4 Gradient from an Equation ο£ Find the gradient of a straight line graph from its equation. π¦ = 1 − 2π₯ Putting in form π = ππ + π: π = −ππ + π Gradient is -2 ? 2π₯ + 3π¦ = 4 Putting in form π = ππ + π: ππ = −ππ + π π ? π π²=− π+ π π Gradient is − π π Gradient using two points ! Given two points on a line, the gradient is: πβππππ ππ π¦ (π ππ π) π= πβππππ ππ π₯ (π π’π) 1, 4 5, 7 2, 2 (3, 10) π = 3? (8, 1) ? π = −2 (−1, 10) 8 π = −? 3 Exercise 1 1 2 Determine the gradient of the lines which go through the following points. a 3,5 , 5,11 b −1,0 , 4,3 c 2,6 , 5, −3 d 4,7 , 8,10 e f g h 1,1 , −2,4 3,3 , 4,3 4, −2 , 2, −4 −3,4 , 4,3 π = π? π π= ? π π = −π ? π π= ? π π = −π ? π = π? π = π? π π = −? π Determine the gradient of the lines with the following equations: a π¦ = 5π₯ − 1 π = π? π = −π b π₯+π¦=2 ? π=π c π¦ − 2π₯ = 3 ? d π₯ − 3π¦ = 5 e 2π₯ + 4π¦ = 5 f 2π¦ − π₯ = 1 g 2π₯ = 3π¦ − 7 3* π π= ? π π π = −? π π π= ? π π π= ? π A line π1 goes through the points (2,3) and 4,6 . Line π2 has the equation 4π¦ − 5π₯ = 1. Which has the greater gradient: π π ππ = ππ = π ? π So ππ has greater gradient. Gradients on a curved graph • You have already seen how to calculate the gradient of a linear graph by dividing the change in y by the change in x • Also known as ‘rise over run’ • What if you were asked to calculate the gradient of this line? ο The problem on a curved line is that the gradient is different depending on where on the line you are… ο On the left side, the gradient will be negative ο On the right side, the gradient will be positive Line is downward sloping so the gradient is negative Line is upward sloping so the gradient is positive ο However, we can use a technique to estimate the gradient, using what you already know… Gradients on a curved graph • Estimate the gradient of this line where x = 1 Draw a straight line that just touches the curve and equal space either side of where x = 1 is. ο This line is known as a tangent to the curve 4 ο You can calculate the gradient of it like on a straight line graph ο The value will be an estimate of the gradient of the curve at the given point 2 πΊπππππππ‘ = πππ π ππ’π πΊπππππππ‘ = 4 2 πΊπππππππ‘ = 2 Gradient of a Cubic Graph f(x) = x3 + 3x2 + 2x Find the gradient where x = -1 πΊπππππππ‘ = πππ π ππ’π −4 πΊπππππππ‘ = 4 4 πΊπππππππ‘ = −1 4 Area under a Graph Speed-Time graphs How would you find the distance travelled in this case? Speed (m/s) width length Time (s) Key Points: The object is moving with a constant speed. The area under this graph would be a rectangle. Area = length x width Speed-Time graphs [Task 1] Speed (m/s) One I prepared earlier! [Task 2] Mr Carne’s Max Speed was 1.8 m/s Time (s) Area = ½ x Base x Height How far did Mr Carne travel in the first 6 seconds? Area = ½ x 6 x 1.8 Shape under the graph for the first 6 seconds is a triangle. The area of this equals the distance travelled in the first 6 seconds. Area = 5.4 m Speed-Time graphs Speed (m/s) One I prepared earlier! Time (s) Area = length x width How far did Mr Carne travel when at a constant speed? Area = 1.8 x 6 Shape under the graph for a constant speed is a rectangle. The area of this equals the distance travelled while travelling at a constant speed. Area = 10.8 m Speed-Time graphs Speed (m/s) One I prepared earlier! Time (s) We have calculated this. We have calculated this. What is the total distance travelled by Mr Carne? The total distance travelled by Hamilton is equal to the whole area underneath the graph. We must now calculate this and total all three areas together. Speed-Time graphs Speed (m/s) Another one I prepared earlier! Time (s) Area = ½ x Base x Height How far did Mr Carne travel between 12 and 15 seconds? Shape under the graph for the this period of time is a triangle. The area of this equals the distance travelled between 12 and 15 seconds. Area = ½ x 3 x 1.8 Area = 2.7 m Speed-Time graphs Area = 5.4 m Area = 10.8 m Area = 2.7 m Speed (m/s) Another one I prepared earlier! Time (s) Total distance travelled = Total area under the graph Total distance travelled = 5.4 + 10.8 + 2.7 Total distance travelled = 18.9 m