Linear graphs and models UNIT ONE GENERAL MATHS

INTRODUCTION TO LINEAR GRAPHS 𝒚=𝟐𝒙+𝟏𝟎 𝟒=𝒚−𝒙 A linear function or equation is a set of ordered pairs that, when graphed, form a straight line. What does a Linear Graph look like? 𝑦=𝑚𝑥+𝑐

Determining the equation of a straight line Now that we know how to find the gradient, we use this to find the equation of the line. We also need to know the point where the line crosses the y-axis (where 𝑥=0). The equation of a straight line is given by: 𝑦=𝑚𝑥+𝑐 y-intercept Where the line crosses the y-axis At this point, 𝑥=0 or (0,𝑦) gradient The ‘steepness’ of the line

Sketching a straight line The Gradient-Intercept Method * Used for graphing in the form 𝑦=𝑚𝑥+𝑐 * So, the y-intercept is immediately known (0,𝑦), And we find the other point using 𝑚 (the gradient) The 𝑥−𝑦 Intercept Method * Used for graphing in the form 𝑎𝑥+𝑏𝑦=𝑐 𝑜𝑟 𝑎𝑥+𝑏𝑦+𝑐=0 * We find 2 points by solving the equation 1) When 𝑥=0 (0, 𝑦) and when 2) When 𝑦=0 (𝑥,0)

Sketching a straight line THE GRADIENT – INTERCEPT METHOD * Used for graphing in the form 𝑦=𝑚𝑥+𝑐 form. * So, the y-intercept is immediately known (0,𝑦), And we find the other point using 𝑚 (gradient) eg1. Use the gradient-intercept method to graph 𝒚= 𝟑 𝟐 𝒙+𝟏 From this equation, we can see that the y-intercept is 1, which gives point (0, 1) To sketch, we need one more point. We can find this from gradient, 𝑚= 3 2 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 This tells us, for every 3 points up (risen), we move across 2 points (run). (2, 4) (0, 1)

Sketching a straight line THE GRADIENT – INTERCEPT METHOD eg2. Use the gradient-intercept method to graph 𝒚=𝟐𝒙 From this equation, we can see that the y-intercept is 0, which gives point (0, 0) To sketch, we need one more point. We can find this from gradient, 𝑚=2= 2 1 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 This tells us, for every 2 points up (risen), we move across 1 point (run). (3, 6 ) (2, 4 ) (1, 2 ) (0, 0) (-1, -2 ) We could plot more points in this manner…..

Sketching a straight line THE GRADIENT – INTERCEPT METHOD eg3. Use the gradient-intercept method to graph 𝒚=𝟒𝒙+𝟐 From this equation, we can see that the y-intercept is 2, which gives point (0, 2) To sketch, we need one more point. We can find this from gradient, 𝑚=4= 4 1 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 This tells us, for every 4 points up (risen), we move across 1 point (run). (1, 6) (0, 2)

Sketching a straight line The Gradient-Intercept Method * Used for graphing in the form 𝑦=𝑚𝑥+𝑐 * So, the y-intercept is immediately known (0,𝑦), And we find the other point using 𝑚 (the gradient) The 𝑥−𝑦 Intercept Method * Used for graphing in the form 𝑎𝑥+𝑏𝑦=𝑐 𝑜𝑟 𝑎𝑥+𝑏𝑦+𝑐=0 * We find 2 points by solving the equation 1) When 𝑥=0 (0, 𝑦) and when 2) When 𝑦=0 (𝑥,0)

Sketching a straight line THE 𝑥−𝑦 INTERCEPT METHOD * Good for graphing in the form 𝑎𝑥+𝑏𝑦=𝑐 𝑜𝑟 𝑎𝑥+𝑏𝑦+𝑐=0 * We find 2 points by solving the equation 1) When 𝑥=0 (0, 𝑦) and 2) When 𝑦=0 (𝑥,0) eg1. Sketch the line given by the equation 2𝑥+3𝑦=6 Let 𝑥=0: 2 0 +3𝑦=6 3𝑦=6 𝑦=2 (0, 2) Let y=0: 2𝑥+3(0)=6 2𝑥=6 𝑥=3 (3, 0) (0, 2) (3, 0)

Sketching a straight line THE 𝑥−𝑦 INTERCEPT METHOD * We find 2 points by solving the equation 1) When 𝑥=0 (0, 𝑦) and 2) When 𝑦=0 (𝑥,0) eg2. Sketch the line given by the equation −4𝑥+2𝑦=8 Let 𝑥=0: −4 0 +2𝑦=8 2𝑦=8 𝑦=4 (0, 4) Let y=0: −4𝑥+2(0)=8 −4𝑥=8 𝑥=−2 (−2, 0) (0, 4) (-2, 0)

Sketching a straight line THE 𝑥−𝑦 INTERCEPT METHOD * We find 2 points by solving the equation 1) When 𝑥=0 (0, 𝑦) and 2) When 𝑦=0 (𝑥,0) eg3. Sketch the line given by the equation 3𝑥+𝑦−9=0 Let 𝑥=0: 3 0 +𝑦−9=0 𝑦−9=0 𝑦=9 (0, 9) Let y=0: 3𝑥+ 0 −9=0 3𝑥−9=0 3𝑥=9 𝑠𝑜 𝑥=3 (3, 0) (0, 9) (3, 0)

Sketching a straight line THE 𝑥−𝑦 INTERCEPT METHOD * We find 2 points by solving the equation 1) When 𝑥=0 (0, 𝑦) and 2) When 𝑦=0 (𝑥,0) eg4. Sketch the line given by the equation 𝑦=3𝑥+4 Let 𝑥=0: 𝑦= 3×0 +4 𝑦=4 Let y=0: 0=3𝑥+4 −4=3𝑥 𝑥= −4 3 (0, 4) 0, 4 ( −4 3 , 0) ( −𝟒 𝟑 , 0)

NOW DO – WORKSHEET ONE

SKECTHING A STRAIGHT LINE USING YOUR CALCULATOR TO PLOT A GRAPH Go to MENU then choose Lets try graphing 𝒚=−𝟒𝒙+𝟔 * Then enter the equation to graph into “y1” * Highlight the equation and drag it into the graph. * The x – intercept can be found using: Analysis  g-solve  root * The y – intercept can be found using: Analysis  g-solve  y-intercept

SKECTHING A STRAIGHT LINE USING YOUR CALCULATOR TO PLOT 2 GRAPHS AND FIND POINT OF INTERSECTION Go to MENU then choose Lets try 𝒚=−𝟒𝒙+𝟔 and 𝒚=𝒙+𝟏 Enter one equation into y1, the other to y2. Highlight and drag each into the graph.

SKECTHING A STRAIGHT LINE USING YOUR CALCULATOR TO PLOT 2 GRAPHS AND FIND POINT OF INTERSECTION Find the intersection of 𝒚=−𝟒𝒙+𝟔 and 𝒚=𝒙+𝟏 So the point of intersection is (1, 2)

Now – check your answers from worksheet one with the classpad

Determining the equation of a straight line Now that we know how to find the gradient, we use this to find the equation of the line. We also need to know the point where the line crosses the y-axis (where 𝑥=0). The equation of a straight line is given by: 𝑦=𝑚𝑥+𝑐 y-intercept Where the line crosses the y-axis At this point, 𝑥=0 or (0,𝑦) gradient The ‘steepness’ of the line

The gradient of the graph

Determining the equation of a straight line 𝑦=𝑚𝑥+𝑐 ① Write the equation for a line which crosses the y-axis at 𝑦=1 and has a gradient of 6 ② Write the equation for a line which crosses the y-axis at 𝑦=−2 and has a gradient 3 ③ Write the equation for a line with a gradient of -2, and passes through the point (0, 1) ④ Write the equation for a line which crosses the y-axis at 𝑦=0 and has a gradient of ½ 𝒚=𝟔𝒙+𝟏 𝒚=𝟑𝒙 −𝟐 𝒚=−𝟐𝒙+𝟏 𝒚=𝟎.𝟓𝒙

Determining the equation of a straight line 𝑦=𝑚𝑥+𝑐 Find the equation for the graph below, which has gradient of 4. 𝒚=𝟒𝒙 −𝟑

Determining the gradient and y-intercept State the gradient and y-intercept of the following equations: ① 𝑦=5𝑥+2 ② 𝑦= 3𝑥 + 6 2 ③ 𝑦=7−3𝑥 ④ 𝑦= 5 + 8𝑥 2 Gradient = 5 y-intercept = 2 Gradient = 𝟑 𝟐 y-intercept = 3 Gradient = -3 y-intercept = 7 Gradient = 𝟒 y-intercept = 2.5

Determining the gradient and y-intercept State the gradient and y-intercept of the following equations: ① 2𝑦=6𝑥+2 ② 6x+3𝑦=9 ③ 4𝑦=16−12𝑥 ④ 2𝑦−3𝑥=10 → 𝑦=3𝑥+1 Gradient = 3 y-intercept = 1 → 3𝑦=−6𝑥+9 → 𝑦=−2𝑥+3 Gradient = -2 y-intercept = 3 → 𝑦=4−3𝑥 Gradient = -3 y-intercept = 4 → 2𝑦=3𝑥+10 → 𝑦=1.5𝑥+5 Gradient = 1.5 y-intercept = 5

Determining the gradient given 2 points If we are given the coordinates on two points on the graph ( 𝑥 1 , 𝑦 1 ) and ( 𝑥 2 , 𝑦 2 ) we can find the gradient using the following equation: 𝑚=𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 eg. Find the gradient of the line that includes the points (3, 6) and (5, 12) 𝑚=𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 12−6 5−3 = 6 2 =3

Determining the gradient given 2 points 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡=𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) eg2. Find the gradient of the line graphed: 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) = (3−0) (0−3) = 3 −3 =−1 𝑜𝑟 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) = (0−3) (3−0) = −3 3 =−1 Using the two points given on our graph, we can substitute these into the gradient ‘m’ equation above. It doesn’t matter which point we call ( 𝑥 1 , 𝑦 1 ) and which we call ( 𝑥 2 , 𝑦 2 ), as either will give us the same result. PROOF :

Determining the gradient given 2 points 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡=𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) eg3. Find the gradient of the line graphed: So, using points given (-2, 0) and (0, 6) 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) = (6−0) (0−−2) = 6 2 =3 𝑜𝑟 𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) = (0−6) (−2−0) = −6 −2 =3

Determining the gradient given 2 points 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡=𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) eg4. A line has a gradient of 2 and passes through the points 2, 5 and (𝑎, 13) . Find the value of a. 𝑚= ( 𝑦 2 − 𝑦 1 ) ( 𝑥 2 − 𝑥 1 ) 2= 13−5 𝑎−2 2 𝑎−2 =8 2𝑎−4=8 2𝑎=12 𝑎= 12 2 𝑎=6

The gradient of the graph

Now Do – exercise 10.2 Q1,2,3,5,6,9,10ac,12,13,16,17,19,20,21,24

Determining the equation of a line We can find the equation of a straight line graph if we are given the gradient and any point on the line. Step One: Substitute the gradient ‘m’ into the equation 𝑦=𝑚𝑥+𝑐 Step Two: Now choose any point on the line and sub it’s coordinates into your equation from Part One. This leaves you with an unknown ‘c’. Solve for ‘c’ Step Three: Sub your found ‘m’ and ‘c’ value into the equation 𝑦=𝑚𝑥+𝑐 This is your equation for the straight line.

Determining the equation of a line We can find the equation of a straight line graph if we are given the gradient and any point on the line. Example One: Write the equation for a line with a gradient of 3, and passes through the point (2, 7) 𝑦=𝑚𝑥+𝑐 𝑦=3𝑥+𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑚 𝑣𝑎𝑙𝑢𝑒) 7=3 2 +𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑥,𝑦 𝑣𝑎𝑙𝑢𝑒𝑠) 7=6+𝑐 (𝑠𝑜𝑙𝑣𝑒 for c) 1=𝑐 (𝑢𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑦=𝑚𝑥+𝑐) So the equation for the line is: 𝒚=𝟑𝒙+𝟏

Determining the equation of a straight line 𝑦=𝑚𝑥+𝑐 Example Two: Write the equation for a line with a gradient of 2, and passes through the point (3, 1) y=mx+c 𝑦=2𝑥+𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑚 𝑣𝑎𝑙𝑢𝑒) 1=2 3 +𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑥,𝑦 𝑣𝑎𝑙𝑢𝑒𝑠) 1=6+𝑐 (𝑠𝑜𝑙𝑣𝑒 for c) −5=𝑐 (𝑢𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑦=𝑚𝑥+𝑐) So the equation for the line is: 𝒚=𝟐𝒙−𝟓

finding the equation of a line given 2 points We can also find the equation of a straight line graph if we are given any 2 points on the line. Step One: Find the gradient using 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 Step Two: Substitute ‘m’ into the equation 𝑦=𝑚𝑥+𝑐 Step Three: Now choose one of the points and sub it’s coordinates into your equation from Part Two. This leaves you with an unknown ‘c’. Solve for ‘c’ Step Four: Sub your found ‘m’ and ‘c’ value into the equation 𝑦=𝑚𝑥+𝑐 This is your equation for the straight line.

finding the equation of a line given 2 points Example One: Write the equation for a line that passes through the points (1, 7) and 4, 13 Find m: 𝑚= 13−7 4−1 = 6 3 =2 𝑦=𝑚𝑥+𝑐 𝑦=2𝑥+𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑚 𝑣𝑎𝑙𝑢𝑒) 7=2 1 +𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑥,𝑦 𝑣𝑎𝑙𝑢𝑒𝑠) 7=2+𝑐 (𝑠𝑜𝑙𝑣𝑒 for c) 5=𝑐 (𝑢𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑦=𝑚𝑥+𝑐) So the equation for the line is: 𝒚=𝟐𝒙+𝟓

finding the equation of a line given 2 points Example Two: Write the equation for a line that passes through the points (−2, 10) and 2, 18 Find m: 𝑚= 18−10 2−−2 = 8 4 =2 𝑦=𝑚𝑥+𝑐 𝑦=2𝑥+𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑚 𝑣𝑎𝑙𝑢𝑒) 10=2 −2 +𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑥,𝑦 𝑣𝑎𝑙𝑢𝑒𝑠) 10=−4+𝑐 (𝑠𝑜𝑙𝑣𝑒 for c) 14=𝑐 (𝑢𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑦=𝑚𝑥+𝑐) So the equation for the line is: 𝒚=𝟐𝒙+𝟏𝟒

Displaying data on a scatterplot Bivariate Data can be displayed on a scatterplot. What does this look like? Two sets of data that is related is given on a table. This data is used as sets of coordinates. We plot these on an axis and use it to model relationships between variables.

Example: Plot the following data on a scatterplot. Age (years) 1 2 3 4 5 6 7 8 9 10 11 Foot length (cm) 13 15 17 18 20 19 22 24 25

Drawing a line of best fit on a scatterplot Once the data is plotted on the graph, we can add a line of best fit to model the data. How to add a line of best fit Fit the line ‘by eye’ by balancing the data points on either side of the line.

Drawing a line of best fit on a scatterplot How to add a line of best fit Fit the line ‘by eye’ by balancing the data points on either side of the line.

finding the equation of a line of best fit As we did before, we can find the equation of any straight line graph by finding any 2 points on the line. Step One: Find the gradient using 𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 Step Two: Substitute ‘m’ into the equation 𝑦=𝑚𝑥+𝑐 Step Three: Now choose one of the points and sub it’s coordinates into your equation from Part Two. This leaves you with an unknown ‘c’. Solve for ‘c’ Step Four: Sub your found ‘m’ and ‘c’ value into the equation 𝑦=𝑚𝑥+𝑐 This is your equation for the straight line.

finding the equation of a line of best fit Example: Find the equation of the line of best fit. Choose 2 points on the line Find m: 𝑚= 85−65 4−1.5 = 20 2.5 =8 (1.5, 65) (4, 85) 𝑦=𝑚𝑥+𝑐 𝑦=8𝑥+𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑚 𝑣𝑎𝑙𝑢𝑒) 85=8 4 +𝑐 (𝑠𝑢𝑏 𝑖𝑛 𝑡ℎ𝑒 𝑥,𝑦 𝑣𝑎𝑙𝑢𝑒𝑠) 85=32+𝑐 (𝑠𝑜𝑙𝑣𝑒 for c) 53=𝑐 (𝑢𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑤𝑟𝑖𝑡𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑦=𝑚𝑥+𝑐) So the equation for the line is: 𝒚=𝟖𝒙+𝟓𝟑

Using the line of best fit to make predictions Once we have a line of best fit, we can use this to make predictions. We can make predictions using either the graphed line – reading off the graph; or by substituting values into the linear equation and solving. Example: Estimate how many hours a student worked to obtain a grade of 80 Reading off the graph = 3.5 hours

Using the line of best fit to make predictions Once we have a line of best fit, we can use this to make predictions. We can make predictions using either the graphed line – reading off the graph; or by substituting values into the linear equation and solving. Example: Estimate how many hours a student worked to obtain a grade of 80 OR Using the formula from the last slide 𝒚=𝟖𝒙+𝟓𝟑 → 𝑮𝒓𝒂𝒅𝒆= 𝟖 × 𝑯𝒐𝒖𝒓𝒔 𝒘𝒐𝒓𝒌𝒆𝒅 +𝟓𝟑 𝟖𝟎=𝟖𝒙+𝟓𝟑 𝟐𝟕=𝟖𝒙 𝑯𝒐𝒖𝒓𝒔 𝒘𝒐𝒓𝒌𝒆𝒅= 𝟐𝟕 𝟖 =𝟑.𝟒

Using the line of best fit to make predictions Example: Estimate the grade received after completing 2 hours of work. Reading off graph…… Approx 70 OR Using the formula 𝒚=𝟖𝒙+𝟓𝟑 → 𝑮𝒓𝒂𝒅𝒆= 𝟖 × 𝑯𝒐𝒖𝒓𝒔 𝒘𝒐𝒓𝒌𝒆𝒅 +𝟓𝟑 𝐆𝐫𝐚𝐝𝐞=𝟖(𝟐)+𝟓𝟑 𝑮𝒓𝒂𝒅𝒆=𝟏𝟔+𝟓𝟑 𝑮𝒓𝒂𝒅𝒆=𝟔𝟗

Reliability of predictions For the graphed data shown right, the data is plotted and shown in the range between the dotted lines. When making predictions, estimates that fall in this range is called INTERPOLATION – these predictions are reliable, as the line is based on data in this range. Any predictions made relating to data outside of this range is called EXTRAPOLATION – these predictions are not reliable, as the line is based on data outside of this range.

When making predictions, estimates that fall in this range is INTERPOLATION – these predictions are reliable. Any predictions made relating to data outside of this range is EXTRAPOLATION – these predictions are not reliable. eg1. If I use the line to predict what test score someone got in Physics, based on a score of 60 in Maths – would this be a reliable prediction? YES – Interpolation eg2. If I use the line to predict what test score someone got in Physics, based on a score of 25 in Maths – would this be a reliable prediction? NO – Extrapolation

Now Do – exercise 10.4 Q1, 2, 3, 5, 6, 8, 10, 11, 13, 14,15a, 16, 17, 18

Linear modelling Practical problems where there is a constant change over time can be modelled by linear equations. They can be used for problems such as: The hourly Rate charged to complete work. Something being filled up or emptied at a constant flow. We model these scenarios using the straight line equation 𝒚=𝒎𝒙+𝒄 The x-value usually represents 𝑡𝑖𝑚𝑒 and the y-value being the 𝑐ℎ𝑎𝑛𝑔𝑖𝑛𝑔 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 (eg. Total cost, Total amount of water etc..)

Forming linear equations Eg1: An electrician charges $70 per hour plus a weekend callout fee of $80. Form a linear equation to find the total cost, 𝐶, for his services for a period of hours, ℎ, over the weekend. 𝑦=𝑚𝑥+𝑐 𝐶=70ℎ+80

Forming linear equations Eg2: A pool initially has 50 litres of water in it, before it is then filled at a constant rate of 20 litres per minute. Form a linear equation to find the total water in the pool, 𝑊, for every minute, t, that its being filled. 𝑦=𝑚𝑥+𝑐 𝑊=20𝑡+50

Forming linear equations Eg3: A silo holds grain and is filled with 1200kg of grain, before it is then emptied at a constant rate of 30kg per minute. Form a linear equation to find the total grain in the silo, 𝐺, for every minute, t, that its being emptied. 𝑦=𝑚𝑥+𝑐 𝐺=−30𝑡+1200

Forming linear equations Eg4: A person walks at a constant speed of 6km/hr, over a distance, 𝑑. Form a linear equation to find the total distance travelled, 𝑑, for every hour, t, of the walk. 𝑦=𝑚𝑥+𝑐 𝑑=6𝑡

Solving practical problems Now that we can form equations, we can use these to determine how much money/how full/etc.. something is….. Eg1. An electrician charges $70 per hour plus a weekend callout fee of $80. Gives the equation: 𝐶=70ℎ+80 How much does Sunday job cost if he is hired for 3 hours? 𝐶 = 70ℎ + 80 𝐶= 70×3 +80 𝐶=210+80 𝐶=$290

Solving practical problems Eg2. A pool initially has 50 litres of water in it, before it is then filled at a constant rate of 20 litres per minute. Gives the equation: 𝑊=20𝑚+50 a) How much water is in the pool after 30 minutes? 𝑊=20𝑚+50 𝑊= 20×30 +50 𝑊=600+50 𝐶=650 litres

Solving practical problems Eg2. A pool initially has 50 litres of water in it, before it is then filled at a constant rate of 20 litres per minute. Gives the equation: 𝑊=20𝑚+50 b) How much time has passed if the pool has 3000 Litres in it? Give your answer to the nearest minute. 𝑊=20𝑚+50 3000=20𝑚+50 2950=20𝑚 𝑚=147.5 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑚=148 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

3000 – The amount of water initially in the tank. Eg3: A water tank holding 3000 L of water is being drained at a constant rate of 100 Litres per minute. a) Form a linear equation to find the total water in the tank, 𝑊, for every minute, t, that its being emptied. 𝑊=−100𝑡+3000 What is the y-intercept and what does it represent? c) How much water is left in the tank after 15 minutes? 𝑊= −100×15 +3000 =−1500+3000 =1500 𝑙𝑖𝑡𝑟𝑒𝑠 How long does it take for the tank to be emptied? 3000 – The amount of water initially in the tank. 0=−100𝑡+3000 −3000=−100𝑡 𝑡= −3000 −100 =30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

Now Do – exercise 10.3 Q1, 2, 3, 7, 10, 11, 13, 14, 17

The domain of a linear model In practice, there are usually limitations to problems. In maths, we can define these limits – this is called the domain. Eg1. A long distance runner runs at a constant speed of 12km/hr for 3 hours, over a distance, 𝑑. Form a linear equation to find the total distance travelled, 𝑑, for every hour, t, of the run and give the domain of the model. 𝑑=12t , 0≤𝑡≤3

The domain of a linear model Eg2. Julie works part-time at the cinema for $18.20 per hour. She does a minimum of 10 hours a week, to a maximum of 16 hours per week. Form a linear equation to find the total pay, 𝑃, for every hour, t, of the working week and give the domain of the model. 𝑃=18.20t , 10≤𝑡≤16

The domain of a linear model Eg3. Officeworks has a deal on colour printing for 10 cents per page if a minimum of 100 pages are printed, to a maximum of 500 pages. Set Up a linear model for the cost of the printing, including a domain. 𝐶=0.10p , 100≤𝑝≤500

Now Do – exercise 10.3 Q1, 2, 3, 7, 10, 11, 13, 14, 17 plus 5a, 6, 8

Understanding Inequalities What do the following symbols represent? < > ≤ ≥ 𝑥<4 𝑥 𝑖𝑠 4 𝑥>12 𝑥 𝑖𝑠 12 𝑥≤ 8 𝑥 𝑖𝑠 8 𝑥≥−3 𝑥 𝑖𝑠 −3 less than greater than less than or equal to greater than or equal to less than greater than less than or equal to greater than or equal to

Setting up domains – Understanding Inequalities How can we represent these values on a number line? 𝑥>3 𝑥<4 1<𝑥<7 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

Setting up domains – Understanding Inequalities How can we represent these values on a number line? 𝑥≤0 1 ≤𝑥<8 −4<𝑥≤2 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

(the point is not included) Step graphs Are formed by two or more linear graphs that have zero gradients. So they are graphs of equations in the form: 𝑦="𝐴 𝑁𝑢𝑚𝑏𝑒𝑟" Have breaks between graphs, as shown in the diagram The end points of each line will include either a solid dot at the end or an open dot at the end – this depends on whether the point is included in the interval. This can be determined by looking at the domain given. ≥ 𝑜𝑟 ≤ Solid dot at end of line (the point is included) > 𝑜𝑟 < Open dot at end of line (the point is not included)

Step graphs eg. Draw the following graphs as a step graph: 𝑦=1, −4≤𝑥<3 𝑦=5, 3≤𝑥≤9

Step graphs eg2. Draw the following graphs as a step graph: 𝑦=−5, −8<𝑥≤0 𝑦=1, 0≤𝑥≤4 𝑦=8, 4<𝑥≤10

Step graphs eg3. Write the equations that make up the following step graph 𝑦=6, −8≤𝑥≤−4 𝑦=−4, −4<𝑥≤2 𝑦=3, 2<𝑥≤10

Step graphs eg4. Write the equations that make up the following step graph 𝑦=−3, −7≤𝑥≤−1 𝑦=0 −1<𝑥≤4 𝑦=6, 4<𝑥≤10

eg5. The cost to hire rollerskates on the beach is listed on the table below. Construct a step graph to model the cost for up to 6 hours. Time (hours) Total Cost 0 - 2 $10 2 – 5 $15 5 – 8 $20

eg6. The same hire company rents skateboards, with their pricing structure given on the graph below. Represent this information on a table. Time (hours) Total Cost 0<𝑡≤1 $15 1<𝑡≤2.5 $25 2.5<𝑡≤4 $40 4<𝑡≤5 $45

Now Do – exercise 10.5 :step graphs Q3, 4, 7, 8, 10

Piecewise graphs Are formed by two or more linear graphs that are joined by Points of Intersection. Have no ‘break’ in the graph, the are a continuous graph, as shown in the diagram below. We draw each graph, within a specified domain.

piecewise graphs eg1. Draw the following graphs piecewise graphs: 𝑦=𝑥+1, −6≤𝑥≤3 𝑦=−2𝑥+10, 3≤𝑥≤8 Sketch the graphs on your calculator Use your domains to decide what portion of the graph to draw. HINT: A graphs will end and join the next piece at the Point of Intersection

piecewise graphs eg2. Draw the following graphs piecewise graphs: 𝑦=𝑥+4, 𝑥≤2 𝑦=−2𝑥+10, 𝑥≥2 Sketch the graphs on your calculator Use your domains to decide what portion of the graph to draw. HINT: A graphs will end and join the next piece at the Point of Intersection

piecewise graphs Eg3. Draw the following graphs piecewise graphs: 𝑦=3𝑥, −3≤𝑥≤1 𝑦=3, 1≤𝑥≤6 𝑦=−2𝑥+15, 6≤𝑥≤10 Sketch the graphs on your calculator Use your domains to decide what portion of the graph to draw. HINT: A graphs will end and join the next piece at the Point of Intersection

Now Do – linear piecewise worksheet 1 q1,2,3

Piecewise graphs – finding the domain values We are not always told what the endpoints of each part of the graph are, instead we have to find the endpoints before we can draw the graph. We do this by solving the equations simultaneously – ie. finding the POINT OF INTERSECTION of each graph. The ‘x’ value of the P.O.I will be the endpoint (the domain) of your graph. Eg1. Given the graph below, complete the blank spaces in the domain, by solving the equations simultaneously. 𝑦=−1.5𝑥+1.5, 𝑥≤ ⎕ 𝑦=3, ⎕≤𝑥≤⎕ 𝑦= 1 3 𝑥+1, 𝑥≥⎕ 1 1 6 6

eg2. A piecewise graph is constructed from the following graphs: 𝑦=3𝑥+2, 𝑥≤𝑎 𝑦=𝑥+4, 𝑥≥𝑎 By solving the equations simultaneously, find the value of 𝑎. Sketch using the Calculator and find the P.O.I. The value of ‘x’ in your P.O.I is a. 𝑎 = 1 b) Now sketch the piecewise graph given their domains.

eg3. A piecewise graph is constructed from the following graphs: 𝑦=−2𝑥−7, 𝑥≤𝑎 𝑦=3𝑥−2 𝑎≤𝑥≤𝑏 𝑦=−𝑥+10 𝑥≥𝑏 By solving the equations simultaneously, find the value of 𝑎. Sketch the first two equations using your Calculator and find the P.O.I. The value of ‘x’ in your P.O.I is a. Point of Intersection: Analysis  gsolve  Intersection P.O.I at (−1, −5) So: 𝒂 =−𝟏

eg3. A piecewise graph is constructed from the following graphs: 𝑦=−2𝑥−7, 𝑥≤−1 𝑦=3𝑥−2 −1≤𝑥≤𝑏 𝑦=−𝑥+10 𝑥≥𝑏 b) By solving the equations simultaneously, find the value of 𝑏. Sketch the second two equations using your Calculator and find the P.O.I. The value of ‘x’ in your P.O.I is a. Point of Intersection: Analysis  gsolve  Intersection P.O.I at ( 3, 7) So: 𝒃 =𝟑

eg3. A piecewise graph is constructed from the following graphs: 𝑦=−2𝑥−7, 𝑥≤−1 𝑦=3𝑥−2 −1≤𝑥≤3 𝑦=−𝑥+10 𝑥≥3 c) Now sketch the piecewise graphs Sketch all equations using your Calculator Re-draw using the defined domains.

Now Do – linear piecewise worksheet 1 q4 and 5

Piecewise – applications problems Eg. The following equations represent a pool being filled over 50 minutes. It is initially filled at a constant rate, before changing the water flow part way through. Time is given by t (minutes), Total Water is given by W (litres) 𝑊=10𝑡, 0≤𝑡≤𝑎 𝑊=15𝑡−100, 𝑎≤𝑡≤𝑏 Find the value of 𝑎. Sketch the two equations using your Calculator and find the P.O.I. The value of ‘x’ in your P.O.I is a. Point of Intersection: Analysis  gsolve  Intersection 𝑎 = 20 What does this answer mean in the context of the problem?

Piecewise – applications problems Eg. The following equations represent a pool being filled over 50 minutes. It is initially filled at a constant rate, before changing the water flow part way through. Time is given by t (minutes), Total Water is given by W (litres) 𝑊=10𝑡, 0≤𝑡≤20 𝑊=15𝑡−100, 20≤𝑡≤𝑏 b) What is the value of 𝑏 ? 𝑏 =50 As the pool is being filled over 50 minutes.

Piecewise – applications problems Eg. The following equations represent a pool being filled over 50 minutes. It is initially filled at a constant rate, before changing the water flow part way through. Time is given by t (minutes), Total Water is given by W (litres) 𝑊=10𝑡, 0≤𝑡≤20 𝑊=15𝑡−100, 20≤𝑡≤50 c) Sketch the piecewise graph 𝑊 (𝑙𝑖𝑡𝑟𝑒𝑠) (50, 650) (20, 200) 𝑡 (𝑚𝑖𝑛𝑠)

Now Do – linear piecewise worksheet 2 Q1 – 4