Module 3: Constructing and interpreting linear graphs Chapter 16

The gradient of a straight line We already know from the core unit that the equation of a straight line can be stated by finding the gradient and the vertical intercept. The gradient measures the slope of the line. The value of the gradient is the ratio between the rise and the run. The rise is the change in the vertical values between any two points. The run is the change in the horizontal values between the same two points.

Example rise = 8 run = 4 rise 8 Gradient =  =  = 2 run 4 y x 8 4 4 8 6 4 2 2 4 6 8 run = 4 4 4 rise 8 Gradient =  =  = 2 run 4

Formula to calculate the gradient from two points y (x2, y2)  y2  y1 rise = y2  y1 x (x1, y1)  run = x2  x1 x2  x1 rise y2  y1 Gradient =  =  x2  x1 run

Example (4, 4)  e.g., (x1, y1) = (0, 4) and (x2, y2) = (4, 4) To calculate the gradient of the line, choose any two points on the line. y (4, 4)  4 4  (4) e.g., (x1, y1) = (0, 4) x and (x2, y2) = (4, 4) 8 6 4 2 2 4 6 8 y2  y1 Gradient =  x2  x1 (0, 4)  4 4  (4) 4  0 =  4  0 8 =  = 2 4

Example e.g., (x1, y1) = (0, 4) and (x2, y2) = (4, 4) y2  y1 To calculate the gradient of the line, choose any two points on the line. y  (0, 4) 4 8 e.g., (x1, y1) = (0, 4) x and (x2, y2) = (4, 4) 8 6 4 2 2 4 6 8 y2  y1 Gradient =  x2  x1  (4, 4) 4 4  4 4 =  4  0  8 =  = 2 4

Questions Exercise 16A Pages 402 – 403 All

Positive and negative gradients y y (x2, y2) (x2, y2) x x (x1, y1) (x1, y1) Positive gradient: Negative gradient: as x values increase as x values increase y values increase y values decrease

The general equation of a straight line The general equation of a straight line is y = mx + c, where m is the gradient of the line and c is equal to the y-axis intercept. This form, expressing the relation in terms of y, is called the gradient form.

Example Find the gradient and y-axis intercept of the graph of y = 3x − 4. Gradient = 3 Y intercept = -4 y = -2x + 4. y = 6x − 5. 2y = 8x + 6

Sketching the equation of a straight line If we are given the rule of a straight line, we can sketch the graph using the gradient and the y-axis intercept. Mark in the y intercept first Draw the gradient from this point Sketch the graph of y = 3x +1 Sketch the graph of 3y + 6x = 9 Sketch the graph of 3x – y = 6

Parallel Lines Lines that have the same gradient will be parallel.

Questions Exercise 16B Pages 405 All

Finding the equation of a straight line given the gradient and the y intercept y = mx + c A line cuts the y axis at 3 and has a gradient of -2. What is its equation. y = -2x + 3

Finding the equation of a straight line given the gradient and a point. Find the equation of the line that passes through the point ( 3, 2 ) and has a gradient of -4. y = mx + c 2 = -4 * 3 + c 2 = -12 + c c = 14 y = -4x + 14

Finding the equation of a straight line given two points Find the equation of the straight line passing through the points (1, -2) and (3, 2) First, find the gradient rise y2  y1 Gradient =  =  run x2  x1 2 - -2 Gradient =  3 - 1 4 Gradient =  2 Gradient = 2

Finding the equation of a straight line given two points Then use the gradient and one of the points to find the equation m = 2 and ( 3 , 2 ) y = mx + c 2 = 2 * 3 + c 2 = 6 + c c = -4 y = 2x – 4

Finding the equation of a straight line given the graph

Graphs of vertical and horizontal lines All horizontal lines will have an equation of the form y = c where c is the vertical intercept. All vertical lines will have an equation of the form x = k where k is the horizontal intercept

Questions Exercise 16C Page 408 - 409 All

The intercept form of a linear equation Not all equations will be given to you the form y = mx + c An alternative notation is, ax + by = c where a, b & c are constants. The best way to sketch an equation when it is given in this form is to find the x and y intercepts.

Example Sketch the equation of 3x – 2y = 6 3x – 2 * 0 = 6 3x = 6 x = 2 To find the x intercept, let y = 0 3x – 2 * 0 = 6 3x = 6 x = 2 To find the y intercept, let x = 0 3 * 0 – 2y = 6 - 2y = 6 y = -3

Exercises Exercise 16D Page 410 Question 1

Linear Models

Example Construct a rule for the Cost of hiring the taxi. C = 3.20 + 1.60k How much will a journey of 10km cost? $19.20 How far can you travel with $30? 16.75km Sketch the graph of the relationship between cost and the number of kilometres travelled for journeys from 0 to 20 kilometres. What does the vertical intercept represent? What does the gradient represent?

Exercise Exercise 16E Pages 411 – 412 All

Solving Simultaneous Equations by the graphical method Simultaneous equations involves finding the point of intersection of two linear equations at the same time Solve the following two equations simultaneously y = 3x – 4 y = -2x + 1

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method

Solving Simultaneous Equations by the graphical method Solve graphically the simultaneous equations x – y = 5 and x = 2

Solving Simultaneous Equations using the calculator Solve(y = 3x - 4 and y = -2x + 1 ,{x,y})

Using a Calculator to solve Simultaneous Equations x + y = 3 2x – y = -9

Calculator x + y = 3 2x – y = -9

Calculator x + y = 3 2x – y = -9

Calculator 2x - 5y = 11 5x + 3y = 12

Questions Exercise 16F Page 416, as many as you need.

Practical applications of simultaneous equations The perimeter of a rectangle is 48 cm. If the length of the rectangle is three times the width, determine its dimensions. Let w represent the width. Let l represent the length 2l + 2w = 48 l = 3w Use the substitution method to solve.

Practical applications of simultaneous equations Two families went to the theatre. The first family bought tickets for 3 adults and 5 children and paid $73.50 The second family bought tickets for 2 adults and 3 children and paid $46.50 Let a represent the cost of an adult ticket Let c represent the cost of a child ticket. Express these statements as equations. 3 a + 5 c = 73.50 2 a + 3 c = 46.50 Calculator

Questions Exercise 16G Page 417-418 all

Break Even Analysis You decide to go into business making calculators. Each week you find that there are always fixed costs that are independent of the cost of making the actual calculator i.e. wages, rent, insurance, utilities, etc. There are also ongoing costs to buy the parts required for calculator.

Break even analysis You find that the fixed costs of producing the calculator is $1500 and the cost for the parts for each calculator is $30. Represent this relationship as an equation where ‘C’ represents the total cost per week and ‘n’ represents the number of calculators made in that week. C = 1500 + 30n

Break even analysis The calculators are sold which provides revenue for your business. You decide to sell the calculators for $150. Represent this relationship as an equation where ‘R’ represents the revenue and ‘n’ represents the number of calculators sold in that week. R = 150n

Break even analysis The break even point is when the revenue earned is equal to the costs incurred. R = C Before the break even point your business will be making a loss. After the break even point your business will be making a profit

Break even analysis The break even point can be found algebraically or graphically.

Algebraically Revenue = Cost 150n = 30n + 1500 120n = 1500 n = 12.5 The Revenue and Costs at the breakeven point will both be $1875 The business will begin to make a profit when you sell 13 calculators.

Graphically

Graphically C = 1500 + 30n

Graphically R = 150n

The Profit Function The profit made is equal to the Revenue earned less the costs incurred. Profit = Revenue – Costs Profit = 150n – ( 30n + 1500) Profit = 150n – 30n – 1500 Profit = 120n – 1500 Calculate the profit made when you sell 35 calculators. Profit = 120 * 35 – 1500 Profit = $2700

Questions Exercise 16H Page 419 All