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1.3 ALGEBRA Relationships between tables, equations or Graphs 4 Credits (External)

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Note 1: Gradients The gradient of a graph tells us how steep the graph is. It is given as a whole number or a fraction. The top number is the change in y, and the bottom number is the change in x. Gradient =

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Note 1: Gradients Special gradients to remember A horizontal line has a gradient of 0 (the rise = 0) Gradient = 0 A vertical line has a gradient that is undefined. (the run = 0) Gradient = undefined A line that is going up hill is positive. Gradient = positive A line that is going down hill is negative. Gradient = negative

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To draw lines with a certain gradient: 1.) Pick a starting point. 2.) Count the number of squares Up (if rise is +) Down (if rise is -) 3.) Count the number of squares Right (if run is +) Left (if run is -) If there is no run, (i.e. the gradient is a whole number) it is always 1. 4.) Mark the endpoint. Note 1: Gradients

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e.g. Draw lines with the following gradients. a = b = 3 Note 1: Gradients

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To find the gradients of a straight line: Pick 2 points on the line that go through a corner of a grid line. The points marked here are the only ones that can be used as they are the only ones that go through the corners of the grid. Count the rise from one point to the other, then count the run. Simplify where possible. = Note 1: Gradients

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e.g.Find the gradients of the following lines. abcdef b= undefined GAMMA Pg 100 Ex. 8.02 GAMMA Pg 100 Ex. 8.02 = Note 1: Gradients

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Note 2: Graphs of linear functions The graph of a linear function is a straight line. Plot points by: – Draw a table of x values. – Substitute the x value into the formula and solve for y. – Plot the pair of co-ordinates onto the axes In a linear equation, x and y are only raised to the power of 1. (i.e. Not x 2, x 3 etc) There is a common difference between ordered pairs.

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x -20123 y Co- ordinate(-2, )(-1, )(0, )(1, )(2, )(3, ) Drawing up a table of x values. Substitute the x value into the formula and solve for y. Plot the pair of co-ordinates onto the axes eg:Draw the graph of y = 2x – 1 When x = -2y = 2 -2 – 1 = -5 When x = -1y = 2 -1 – 1 = -3 -5-3135 Note 2: Graphs of linear functions

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Starter Identify if the following relations are linear. xY -27 4 01 1-2 2-5 3-8 Diff. xY -22 4 08 116 232 364 Diff. Common difference = -3 Therefore, Linear No Common difference Therefore, Not Linear -3 2 4 8 16 32

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y = 2x – 3 xy = -13 y = x(x + 2) y = 2 Linear y 1 = 2x 1 - 3 Not Linear x = -13y -1 Not Linear y 1 = x 2 + 2x 1 Linear y 1 = 2 Starter Identify if the following relations are linear.

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Note 3: Straight Line Graphs gradient y-int

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e.g. y = 2 x = 4 Note 3: Straight Line Graphs

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e.g. Draw the following straight lines. a)y = b)y = 4x – 2 c)y = x + 1 c)y = -2x x y y = y = -2x y = 4x - 2 y = x + 1 Note 3: Straight Line Graphs

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Note 4: Straight Line Graphs Using Intercept-Intercept Method To find the x-intercept, set y = 0. To find the y-intercept, set x = 0. Join the two intercepts to make a straight line. e.g. Find the x and y intercepts of the line y = -4x + 12 Find y-intercept, set x = 0 y = -4 (0) + 12 y = 12 therefore the y intercept is (0, 12) Find x-intercept, set y = 0 0 = -4x + 12 4x= 12 x = 3 therefore the x intercept is (3, 0)

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Find x-int, Set y = 0 2x – 0 = 6 2x = 6 x= 3 therefore the x-intercept is (3, 0) Find y-int, Set x = 0 2(0) – y = 6 -y = 6 y = -6 therefore the y-intercept is (0, -6) Draw the graph of 2x – y = 6 Note 4: Straight Line Graphs Using Intercept-Intercept Method

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x-intercept is (3, 0) y-intercept is (0, -6)

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The cover up method – a quick way of graphing a straight line in the form ax + by = c -x + 2y = 8 Is to cover up the x term to get the y-intercept -x + 2y = 8 And then cover the y term to get the x-intercept -x + 2y = 8 Then plot the 2 points. i.e. y = 4 (0,4) i.e. x = -8 (-8,0)

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x y (0,4) (-8,0) -x + 2y = 8 IWB Mathematics Ex 13.04 pg 310 IWB Mathematics Ex 13.04 pg 310 HMWK pg 88

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Starter – Graph to Equation y = 2x + 3 y = -1/2 x + 0 y = 1/3x - 5 y = -4x + 4 2 1 3 4

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Note 5: Applications of the Straight Line Graph Many circumstances in nature and life can be graphed using straight line relationships. e.g.De Vinces specializes in pizza. The approximate relationship between x, the # of pizzas they sell daily and y their daily costs, is given by: y = 10x + 50 Draw a graph showing the number of pizzas they sell on the x-axis, and their daily costs on the y-axis. Use an appropriate scale. (Because the numbers are large, it is best to have the y-axis going up in 20s).

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y = 10x + 50. x y Cost $ Number of Pizzas sold 86 4 210 14 12 120 100 80 60 40 20 140

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What are their costs if they sell 8 pizzas? Either read off the graph or substitute into the equation. If their costs are $100, how many pizzas did they sell? Either read off graph or substitute into the equation. Give the y-intercept. What does it represent? Give the gradient. What does it represent? $130 5 50 If no pizzas are sold, it still costs them $50 10 Represents the fact that each pizza they make, costs $10 Number of Pizzas sold Cost $

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Note 6: Rate of Change If two variables (e.g. distance and time) are directly dependent, the graph of this relation is a straight line. We can calculate the rate of change of one variable compared to the other. rate of change = change in dependent variable change in independent variable e.g. = change in distance change in time Time (hrs) Distance (km) The gradient of the graph is also a measure of the rate of change of the dependent variab le

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e.g. The graph below shows the value of a section in Auckland. The rate of increase shows two distinct slopes. Find the rate in increase before and after 2004. Price in $000 Year Value of section in Auckland 280 320 340 1998 Rate of change = Δ price / Δ time = 330 – 280 2004 – 1998 = 50 6 = $8 333/yr 300 360 400 380 200020022004200620082010 Before 2004 After 2004 = 400 – 330 2010 – 2004 = 70 6 = $11 667/yr

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e.g. An engineer studying the expansion of a bar of metal made the following measurements. At 14°C the length of was 41.4 mm, at 22°C the length was 42.0 mm and at 34°C the length was 42.9 mm. Plot the data on a graph & find the rate of change of length compared to temperature. mm °C Expansion of a metal bar 41 42 43 102030 Plot points (14,41.4), (22,42) & (34,42.9) Rate of change = Δ length / Δ temperature = 42.9 – 41.4 34 – 14 = 1.5 20 = 0.075 mm/°C

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Starter This graph shows the distance a car has travelled after leaving Dunedin. a)Find the rate of change for each of the three sections A = B = C = b.) Give the equation that represent section A of the journey D = c.) Using your equation from b) find what time the car is 68 km from Dunedin. Time (pm) 4 5 6 7 8 9 km 20 40 60 80 100 120 140 A B C 100 km/hr 20 km/hr -65 km/hr 100t - 400 4.68 hrs =4.41 IWB Mathematics Ex 13.06 Pg 316 - 321 IWB Mathematics Ex 13.06 Pg 316 - 321

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Starter Draw a distance-time graph to describe the fable of the tortoise and the hare. (assume 3 km race) Time taken (hours) Distance from start (km) 1 32 2 3 14

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Note 7: Simultaneous Equations The point (x,y) where two linear equations cross, gives a unique solution for x and y that satisfies both equations. 10 5 5 15 These 2 linear equations are simultaneously equal at pt (4,7) (4,7) y = - x + 11 y = 1/2 x + 5 7 = - 4 + 11 7 = 1/2 (4) + 5

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Note 7: Simultaneous Equations Two boys go out for lunch. The first orders 2 pies and 1 sandwich for $13.50 and the other orders 1 pie and 2 sandwiches for $12.00. 10 5 5 15 (5,3.50) Let p be the cost of pies and s be the cost of sandwiches pies (p) snd. (s) 1p + 2s = 12.00 (2) 2p + 1s = 13.50 (1)

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Note 7: Simultaneous Equations An animal nursery needs 12 cans of food to feed 2 dogs and 6 cats. The next day they have 4 dogs and 4 cats and need 16 cans of food. 105 -5 5 15 10 (3,1) How many cans does a cat & a dog need? dogs (d) Cats (c) IWB Mathematics Ex 13.05 Pg 313-314 IWB Mathematics Ex 13.05 Pg 313-314 2d + 6c = 12 (2) 4d + 4c = 16 (1) Dogs need 3 cans and cats need 1 can of food.

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Starter (0,2) 4y = 8 y = 2 Subst. y = 2 into either equation 2x + 3(2) = 6 2x = 0 x = 0

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s

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Parabolas Found in nature, motion, structure & art

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Note 8: Quadratic Functions Quadratic functions are algebraic expressions where the highest power of x is x 2 e.g. y = x 2 y = x 2 + 3x – 5 y = (x-2)(x+2) y = (x+2) 2 the graph of these functions are called parabolas.

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y = x 2 vertex (0,0)

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To draw the graph of a parabola select a range of x values substitute into the equation to get a corresponding y-term Plot the points xy =x 2 -3 -2 0 1 2 3 9 4 1 0 1 4 9 y = x 2

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Transformations of the parabola y = x 2 + 7y = x 2 + 4 y = x 2 y = x 2 3 By adding/ subtracting a constant, the parabola shifts up /down. 1.) Vertical Translation y = x 2 ± c

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y = x 2 By adding/ subtracting a constant to the x variable, the parabola shifts left /right. 2.) Horizontal Translation y = (x+k) 2 y = (x3) 2 y = (x+5) 2 y = (x+2) 2

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y = x 2 3.) Combined – Horizontal & Vertical Translation y = (x+k) 2 + c y = (x4) 2 +2 y = (x+3) 2 4 IWB Mathematics Pg 347 Ex. 15.01 IWB Mathematics Pg 347 Ex. 15.01 To get the y-int: Let x = 0

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Starter Write the equations for each of the parabolas drawn below y = (x+6) 2 - 3 y = (x-4) 2 - 5 y = (x-5) 2 + 4

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Note 9: Parabolas (factorised form) A simple way to graph parabolas when the equation is given in factorised form, is to plot the intercepts. e.g. y = (x - 4)(x + 2) Y-int – set x = 0 X-int – set y = 0 y = (0 - 4)(0 + 2) y = -4 x 2 y = -8 0 = (x - 4)(x + 2) x - 4 = 0 or x + 2 = 0 x = 4 x = -2

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Graph of y = (x – 4)(x + 2) x-int y-int Midpoint of parabola x = 1 Subst x = 1 to get vertex y = (1 - 4 )(1 + 2) = -3 x 3 = 9

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Step by step for graphing factorised parabolas 1.) Find y-int by setting x = 0 2.) Find x-int by setting y = 0 3.) Find the axes of symmetry (midpoint between both of the x-int) 4.) Subst. the x-value that the axes of symmetry cuts through and solve for y. This is the vertex.

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Graph the following parabolas a.) y = (x – 3)(x +4)b.) y = x(x - 6) y-int (x = 0) y = -3 x 4 y = 0 = - 12 x-int (y = 0) 0 = (x – 3)(x + 4) 0 = x (x - 6) x – 3 = 0 x + 4 = 0x = 0 x - 6 = 0 x = 3 x = -4 x = 6

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y = (x – 3)(x +4) y = x(x - 6) (3, -9) (-0.5, -12.25) IWB Mathematics Pg 357-9 Ex. 15.04 IWB Mathematics Pg 357-9 Ex. 15.04 HMWK pg 108-109

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Note 10: Parabolas – Vertical Scale Factors If the coefficient of x 2 is negative, then the parabola is inverted (upside down) e.g.y = x 2 + 2 y = (x + 3) 2 y = (x + 3)(x 2) y = (x + 3)(2 x) Different forms of the same equation

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y = x 2 y = 2x 2 y = 3x 2

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y = x 2 y = 1/2x 2 y = x 2 y = 1/4 x 2 y = 1/10 x 2 IWB Mathematics Pg 351 Ex. 15.02 IWB Mathematics Pg 351 Ex. 15.02 HMWK pg 104-107

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Starter B F E D C J I H G A

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Task: Graph the following parabolas y = -(x – 5)(x – 1)y = 2(x – 1) 2 – 2 * Inverted y – intercept = x – intercepts = Midpoint is x = Vertex = (3, ) -5 5, 1 3 4

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Task: Graph the following parabolas y = 2(x – 1) 2 – 2 * Opens upward Vertex = y – intercept = (1, -2) 0 Over 1, up 1 x2 = 2 Over 1, up 3 x2 = 6 Over 1, up 5 x2 = 10

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Graph to Equation in form y=(x+k) 2 +c y =- (x+7) 2 + 2 2 1 y = x 2 - 5

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Graph to Equation in form y=(x+a)(x+b) y = (x+5)(x-1) 1 2 y = (x-2)(x-4) Check your intercepts are correct once you have composed an equation

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Graph to Equation– Vertical scale factor y =-1/3 (x+9)(x+3) 1 2 y = 2(x-4) 2 - 5 Check your intercepts are correct once you have composed an equation y =-1/3 (x+6) 2 + 3 HMWK pg 112-113 IWB Mathematics Pg 357-359 Ex. 15.04 # 6 -16 IWB Mathematics Pg 357-359 Ex. 15.04 # 6 -16

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Starter C E A D B F

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Note 11: Linear Patterns Linear Patterns are sequences of numbers where the difference between successive terms is always the same. e.g. 4, 7, 10, 13, 16, 19, 22……. This sequence has a common difference of _______ +3 Each Term = 3 x Term # + C Term # (x) 1 2 3 4 5 6 7 ……. C is found by substituting ANY term and term # into the formula C = constant term e.g. 4 = 3 x 1 + C C = 4 - 3 C = 1 Rule to find each term = 3x + 1

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e.g. Identify if the following relation are linear. xY -27 4 01 1-2 2-5 3-8 Diff. xY -22 4 08 116 232 364 Diff. Common difference = -3 Therefore, Linear No Common difference Therefore, Not Linear -3 2 4 8 16 32

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e.g. Find the differences, and if the result is linear find the rule. Term #Sequence 116 213 310 47 54 61 Diff. Common difference = -3 Therefore, Linear -3 Each Term = 3 x Term # + C 16 = 3 x 1 + C 16 = 3 + C 19 = C Each Term = 3 x + 19 (X) -3 Check that your formula works !

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Find the differences, and if the result is linear find the rule. Term #Sequence 1-8 2-6 3-4 4-2 50 62 Diff. Common difference = Therefore, 2 2 2 2 Each Term = 2 x Term # + C -8 = 2 x 1 + C -8 = 2 + C -10 = C Each Term = 2 x 10 (X) 2 Check that your formula works ! NOW YOU TRY THIS ONE -8, -6, -4, -2, 0, 2 …… +2+2 HMWK pg 74-82 linear

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Find the differences, and if the result is linear find the rule. Term #Sequence 16 211 316 421 526 631 Diff. Common difference = Therefore, 5 5 5 5 Each Term = 5 x Term # + C 6 = 5 x 1 + C 6 = 5 + C 1 = C Each Term = 5 x + 1 (X) 5 Check that your formula works ! NOW YOU TRY THIS ONE 6, 11, 16, 21, 26, 31 …… 5 linear HMWK pg 74-82 IWB Mathematics Pg 269-273 Ex. 12.02 # 3 -12 IWB Mathematics Pg 269-273 Ex. 12.02 # 3 -12

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Interpret features of graphs in Context a) How much does the Mighty Minibus Co charge per kilometre? b) Charges by Various Vans include a fixed fee for each trip. How much is this fixed fee? c) For which length of trip do both companies charge the same? d) Write the equation for the Various Vans graph. $1.40/km $30 50 km y = 0.8 x + 30 Jim gets quotes from the Mighty Minibus Co and Various Vans for a minibus. Jim can use the graph below to work out the cost of a trip for each rental company.

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Starter F A C E N K J H D P Q B

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Note 12: Quadratic Patterns Quadratic patterns have a constant second order of differences (differences of differences). Term # Sequence 15 211 321 435 553 675 Diff. 6 10 14 18 22 4 4 4 4 The value of the 2 nd order of differences determines our Rule 2nd 1 means a 0.5n 2 2 n 2 3 1.5n 2 4 2n 2 In general 2 nd Diff n 2 2

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Now that we know what the constant for n 2 we can solve for the rest of the equation by substituting the term # into ____________ Term # (n) Sequence2n 2 Adjustment 15 211 321 435 553 675 2n 2 2 8 18 32 50 72 3 3 3 3 3 3 Rule: 2n 2 + 3 HMWK pg 94-100 IWB Mathematics Pg 277-281Ex. 12.03 IWB Mathematics Pg 277-281Ex. 12.03

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NOW YOU TRY 4, 6, 10, 16, 24, 34…. Quadratic patterns have a constant second order of differences (differences of differences). Term # Sequence 14 26 310 416 524 634 Diff. 2 4 6 8 10 2 2 2 2 The value of the 2 nd order of differences determines our Rule 2nd 1 means a 0.5n 2 2 n 2 3 1.5n 2 4 2n 2 In general 2 nd Diff n 2 2

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Now that we know what the constant is for n 2 we can solve for the rest of the equation by substituting the term # into _______________ Term # (n) Sequencen2n2 Adjustment 14 26 310 416 524 634 n2 n2 1 4 9 16 25 36 3 2 1 0 -2 Rule: n 2 – n + 4 HMWK pg 94-100 IWB Mathematics Pg 277-281Ex. 12.03 IWB Mathematics Pg 277-281Ex. 12.03

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Write the rule for the following patterns a.) 5, 8, 11, 14, 17 ….. b.) 6, 12, 20, 30, 42,…... 3n + 2 Term # Seq 16 212 320 430 542 Diff 6 8 10 12 Diff 2 2 2 2nd 1st Term # Seq 16 212 320 430 542 n2n2 1 4 9 16 25 Diff 5 8 11 14 17 n 2 + 3n + 2n2n2 IWB Mathematics Pg 277-281Ex. 12.03 IWB Mathematics Pg 277-281Ex. 12.03 3 3 3 3

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Starter = Find the gradient of the line joining the points D(1,5) and E(5,2) gradient of DE = = =

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Note 13: Exponential Relationships base exponent

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Graph by plotting points x valuey = 2 x -3 -2 0 1 2 3 0.125 0.25 0.5 1 2 4 8 y = 2 x

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y=2 x +2 the exponential function can be vertically translated up or down by adding or subtracting a constant HMWK pg 119-120 IWB Mathematics Pg 367-369 Ex. 15.07 IWB Mathematics Pg 367-369 Ex. 15.07

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Starter Write the rule for the following sequences 4, 8, 12, 16, 20, 24, …… 10, 26, 48, 76, 110, 150, …… 4n 1 st Difference 2nd Difference 16 22 28 34 40 6 6 6 6 3n 2 3, 12, 27, 48, 75, ……. Difference 7 14 21 28 35 + 7n

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Note 14: Exponential Patterns Exponential patterns do not have a constant difference between terms Term # Sequence 13 25 39 417 533 665 Diff. 2 4 8 16 32 4÷2 = 2 8÷4 = 2 16÷8 = 2 32÷16 = 2 The ratio of the 1st order of differences determines our Rule Ratio of 1 st 2 means a base 2 2 x 3 …… 3 x 4 …… 4 x

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Now that we know our rule is of the form 2 x we can solve for the rest by substituting the term # into _____ Term # (n) Sequence 2x2x Adjustment 13 25 39 417 533 665 2x2x 2 4 8 16 32 64 1 1 1 1 1 1 Rule: 2 x + 1 What will the value of term 9 be? 2 9 + 1 = 513 HMWK pg 116-118 IWB Mathematics Pg 286-290 Ex. 12.04 IWB Mathematics Pg 286-290 Ex. 12.04

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Applications to Parabolas A courier company is investigating leasing an old hangar as storage for its vehicles at night. The hangar has a length of 18.5 m and a parabolic cross-section with an equation y = -0.2x(x – 13.5). Each van needs a space 6 m in length, 2.8 m in width and 2.5 m high to allow for access. Height (m) y = -0.2x(x – 13.5) Width (m) a.) What is the maximum height of the hangar? What are the two x-intercepts 18.5 m (0,0) and (13.5, 0) The maximum height occurs at the midpoint (6.75,0) Solve for y given that x = 6.75 y = -0.2(6.75)(6.75 – 13.5) = (-1.35)(-6.75) = 9.1125 m is the maximum height

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Applications to Parabolas A courier company is investigating leasing an old hangar as storage for its vehicles at night. The hangar has a length of 18.5 m and a parabolic cross-section with an equation y = -0.2x(x – 13.5). Each van needs a space 6 m in length, 2.8 m in width and 2.5 m high to allow for access. Height (m) y = -0.2x(x – 13.5) Width (m) b.) Is there room to fit 12 courier vans? Required width = 2.8 x 4 = 11.2 m 18.5 m Required length = 3 x 6 = 18 m Is there sufficient height? 13.5 m – 11.2 m = 2.3 m spare Subst. x = 1.15 to find the height of the hangar at that point y = -0.2(1.15)(1.15 – 13.5) = (-0.23)(-12.35) = 2.8405 m < 13.5 ok < 18.5 ok = 1.15 on each side < 2.5 ok Yes, there is sufficient room for 12 courier vans

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y-intercept = (0,-4.5) Find an equation for the following parabola in the form y = k (x – c)(x – d)

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Applications to Parabolas Jacqui plans to hang shade cloth over a paved area. The shade cloth will hang between two posts 4 metres apart & will fall in the shape of a parabola. At its lowest point it will hang 2.2 m above the paving. The shade cloth is attached at the top of the 3 metre posts. Post Height (m) 3 2 1 Supports 12-20 Width (m) a.) Find the equation that represents the fall of the shade cloth b.) To support the shade cloth, two supports are attached. Find the equation that models either of the two supports.

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Applications of Parabolas a.) Parabola is of the form y = kx 2 + c At its lowest point it will hang 2.2 metres above the paving. Therefore c = 2.2 y = kx 2 + 2.2 Parabola is of the form y = kx 2 + 2.2 A point that satisfies this equation is(2,3)(-2,3) 3 = k(2) 2 + 2.2 0.8 = 4k k = 0.2 y = 0.2x 2 + 2.2 When x = 1 y = 0.2 (1) 2 + 2.2 y = 2.4 (1, 2.4)

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Applications of Parabolas b.) Line is of the form y = mx ± c 2 points that satisfies this equation are (2,2) (1,2.4) This gradient is true for any 2 points on the line y = - 0.4x + 2.8 Gradient is = 0.4 Multiply both sides by (x-2) y – 2 = - 0.4 (x 2) y – 2 = - 0.4x + 0.8 HMWK pg 114-115 IWB Mathematics Pg 434-442 Ex. 17.02 IWB Mathematics Pg 434-442 Ex. 17.02

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Find an equation for the following parabola in the form y = k (x – c) 2 + d y = k (x +6) 2 - 8 Subst (1,4) into eqn 4 = k (1 +6) 2 - 8 4 = k (7) 2 - 8 12 = 49k k = 0.245 y = 0.245 (x +6) 2 - 8

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Find an equation for the following parabola in the form y = k (x – c) 2 + d and then find the y-intercept y = k (x – 3) 2 + 13 Subst (4.22,3.57) into eqn 3.57 = k (4.22 – 3) 2 + 13 k = -6.34 y = -6.34 (x – 3) 2 + 13 3.57 = k (1.22) 2 + 13 -9.43 = 1.4884k To find the y-intercept, set x = 0 y = -6.34 (0 – 3) 2 + 13 y = -6.34 (9) + 13 y = -44.0 (0,-44)

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The following parabola has been translated 4 units up and 6 units to the left. Give the equation of the parabola in its new position and find its new y-intercept. y = k (x-5) 2 + 2 y = -2.2 (x-5) 2 + 2 y = -2.2 (x+1) 2 + 6 y-intercept Set x = 0 y = -2.2 (0+1) 2 + 6 y = 3.8 (0, 3.8) IWB Mathematics Pg 434-442 Ex. 17.02 IWB Mathematics Pg 434-442 Ex. 17.02

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Applications to Parabolas The occupancy rate, as a percentage, over a period of 12 months for a new hotel, is expected to be represented by the graph drawn below. The graph compromises a straight line (which gives the occupancy rate for the first 3 months) and a parabola (which gives the occupancy rate for the remaining 9 months) Write an equation for each part of the graph 1 2 3 4 5 6 7 8 9 10 11 12 months 100 90 80 70 60 50 40 30 20 10 occupancy % First 3 months (linear) gradient = y-intercept = -30/3 = -10 = 70 y = -10x + 70 Last 9 months (quadratic) vertex = (8, 20) y = k(x-8) 2 + 20 Subst (3, 40) to solve for k 40 = k(3-8) 2 + 20 20 = k(-5) 2 k = 0.8 y = 0.8(x-8) 2 + 20 IWB Mathematics Pg 434-442 Ex. 17.02 IWB Mathematics Pg 434-442 Ex. 17.02

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STARTER Applications of Parabolas A civil engineer designs a model of a 6 metre suspension bridge using the equation, where h is the height, in metres, of the hanging cables above the road and d is the distance from the first pillar. a.) What are possible values for d? h = (d – 3) 2 + 7 when d = 0, h = (0 – 3)2 + 7 6 m 1 st pillar b.) Sketch the graph of for appropriate values of d h = (d – 3) 2 + 7 c.) What is the height, h, of the pillars above the road? h = 16 when d = 6, h = (6 – 3)2 + 7 = 16 The pillars are 16 m tall. 0 d 6

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Applications of Parabolas - MERIT A civil engineer designs a model of a 6 metre suspension bridge using the equation, where h is the height, in metres, of the hanging cables above the road and d is the distance from the first pillar. d.) If vertical cables are hung every 0.5 m along the bridge, determine the length of the cable needed: h = (d – 3) 2 + 7 i.) 2 m from the pillar 6 m 1 st pillar h when d = 2, h = (2 – 3) 2 + 7 = 8 metres ii.) 3.5 m from the pillar when d = 3.5, h = (3.5 – 3) 2 + 7 = 7.25 metres

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Applications of Parabolas - MERIT The height of a parachutist above the ground is modelled by the equation where t is the time in minutes after jumping from the airplane. h = -40t 2 + 4t + 10 000 a.) Sketch the graph of h vs. t b.) How long does it take the parachutist to reach the ground? c.) At what height did the parachutist jump from the plane? set h = 0 ( factorise the quadratic ) set t = 0 h = 10 000 m t = 15.86 min or -15.76 min

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Applications of Parabolas - MERIT A wood turner hones out a bowl according to the formula where d is the depth of the bowl and x is the distance in cm from the centre d = 1/3 x 2 – 27 a.) Sketch the graph for appropriate values of x b.) What is the width of the bowl? from -9 to +9 - width is 18 cm -2.5 = 1/3 x 2 – 27 24.5 = 1/3 x 2 c.) What is the distance away from the centre when d = -2.5? 73.5 = x 2 x = ±8.57 cm (3 sf)

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Applications of Parabolas - MERIT A wood turner hones out a bowl according to the formula where d is the depth of the bowl and x is the distance in cm from the centre d = 1/3 x 2 – 27 d.) What is the depth d, 7 cm away from the centre. d = 1/3 (7) 2 – 27 d = -10.7 cm (3 sf) d = 1/3 x 2 – 27

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Applications of Parabolas - MERIT The cross section of a curved skateboard ramp can be modeled by a parabola, given the equation The distance between B and C (the top points on either side of the ramp) is 6 m. y = 0.2 x 2 a.) Calculate the distance from the ground at the highest point (B or C) y = 0.2(3) 2 = 1.8 m x = 3 or -3 b.) When it rains the ramp partially fills with a puddle of water. The puddle is 0.3125 m deep. How wide is the puddle?

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Applications of Parabolas - MERIT The cross section of a curved skateboard ramp can be modeled by a parabola, given the equation The distance between B and C (the top points on either side of the ramp) is 6 m. y = 0.2 x 2 0.3125 = 0.2(x) 2 x 2 = 1.5625 y = 0.3125 b.) When it rains the ramp partially fills with a puddle of water. The puddle is 0.3125 m deep. How wide is the puddle? The width is 2 x 1.25 = 2.5 m (2 sf) x = 1.25

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Applications of Parabolas (last question in assignment) Andrew takes part in a competition to see which of the players could throw a ball over a 5 m fence from the greatest distance. The ball must be on its way down as it clears the fence. He has a couple of trial and the flight of the ball is shown below. If he stood 43 m from the wall would the ball still clear it? Parabola is of the form y = kx 2 + d 0 = k(25) 2 + 15 -15 = 625k k = -0.024 y = kx 2 + 15 (25, 0) is on the parabola y = -0.024x 2 + 15 y = -0.024(x – 25)(x + 25)

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Applications of Parabolas (last question in assignment) Andrew takes part in a competition to see which of the players could throw a ball over a 5 m fence from the greatest distance. The ball must be on its way down as it clears the fence. He has a couple of trial and the flight of the ball is shown below. If he stood 43 m from the wall would the ball still clear it? y = -0.024(0 +18) 2 + 15 y = 7.224 m he has shifted back 18 metres y = -0.024x 2 + 15 The new equation for the shifted parabola is: y = -0.024(x + 18) 2 + 15 y = -0.024(x + 43)(x – 7) 18 m Find y when x = 0

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Mere buys a small tent for her little sister for Christmas. The top part is modelled by a parabola. There are three zips: BE, DE, and EF. The equation of the frame of the top part of the tent is where y is the height in cm and x is the distance from the centre line in cm. a) What is the height of the top of the tent? b) What is the length of the horizontal zip, EF? c) The length of AC is 32 cm. Find the length of zip BE. At x=0, y = 80 cm At y =0, x = 40 cm, EF = 40 cm At x = 16, y = 67.2 cm

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Excellence d) Two poles, PR and QS help keep the tent upright. They are 25 cm from the centre-line of the tent. The bottom part of the tent is also modelled by a parabola. It cuts the vertical axis at y = –20. The length FG = 10 cm. Find the length of the pole PR. Subst x = -25 into To get y = 48.75 cm Subst (40, -10) to solve for k in y = kx 2 – 20 (from part a) k = 1/160 - 48.75 -25 40 Subst x=-25 into y = x 2 – 20 160 y= -16.09 cm The length of the pole PR is 48.75 + 16.09 = 64.84 cm -16.09 IWB Mathematics Pg 434-442 Ex. 17.02 IWB Mathematics Pg 434-442 Ex. 17.02

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Starter Write equations for the lines shown below. a.) ___________ b.) ___________ c.) ____________ y = 5 y = -2x – 7 y = x + 6

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Merit In decorating the hall for the dance streamers are hung across the hall as shown below. The equation for the height, in metres, of the streamer above the floor is: y = (x – 5) 2 + 3. Diagram NOT to scale (a)(i)What is the least distance between the floor and the streamer ? Vertex - 3 metres (ii)What is the height of the ceiling above the floor of the hall ? Set x = 0 or x = 10 y = (0 – 5) 2 + 3 y = 4 metres (iii)What is the width of the hall ? 10 metres

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Excellence As part of the entrance to the dance the organizers are going to make an archway as shown in the plan. The archway will be 250 cm wide and 250 cm above the floor at its highest point. The curved part of the archway can be modeled by a parabola. The top curve has its lowest point 200 cm above the floor. The lower curve is going to be joined to the upper curve by connecting rods AB that are placed 75 cm from either side of the highest point and position the lower curve 200 cm above the floor at point B. Calculate the length of the rods AB. Find an equation that models the curves used to form the archway. B 250 cm 200 cm 75 cm A B A Diagram NOT to scale y = k (x – 125) 2 + 250 subst. (0, 200) to solve for k 200 = k (0 – 125) 2 + 250 -50 = 15625k k = -0.0032 y = -0.0032 (x – 125) 2 + 250

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Excellence Calculate the length of the rods AB. Find an equation that models the curves used to form the archway. BB 250 cm 200 cm 75 cm A A y = -0.0032 (x – 125) 2 + 250 subst. x = 50 to solve for y y= -0.0032 (50 – 125) 2 + 250 50 Pt B has the coordinates ______ 200 = -0.0032 (50 – 125) 2 + c y = 232 pt is (50, 232) (50, 200) c = 218 y = -0.0032 (x – 125) 2 + 218 The differences between y – values at x = 50 is the length of rod AB 232 – 200 = 32 cm 250 – 218 = 32

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Applications to Parabolas In a science experiment, groups of students were firing small rockets which burnt for about 5 seconds and then fell back to earth. They decided to model the path of the rocket by a straight line and half a parabola. Write an equation for each part of the graph 10 20 30 40 50 60 distance (m) 50 45 40 35 30 25 20 15 10 5 height (m) First 45 metres (linear) gradient = y-intercept = 30/45 = 0.67 20 y =.67x + 20 Last 20 metres (quadratic) vertex = (45, 50) y = k(x-45) 2 + 50 Subst (65, 0) to solve for k 0 = k(65-45) 2 + 50 -50 = k(20) 2 k = -0.125 y = -0.125(x-45) 2 + 50

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Applications to Parabolas Calculate the height of the rocket after it has travelled a distance of 55 m 10 20 30 40 50 60 distance (m) 50 45 40 35 30 25 20 15 10 5 height (m) = 37.5 m 35 = -0.125(x – 45) 2 + 50 Subst y = 35 -15 = -0.125(x – 45) 2 x = 55.95 m y = -0.125(55 – 45) 2 + 50 After what distance travelled is the rocket at a height of 35 metres 120 = (x – 45) 2 10.95 = x – 45 y = -0.125(x-45) 2 + 50

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Applications to Parabolas In a science experiment, groups of students were firing small rockets which burnt for about 5 seconds and then fell back to earth. They decided to model the path of the rocket by a straight line and half a parabola. 10 20 30 40 50 60 distance (m) 50 45 40 35 30 25 20 15 10 5 height (m) and x = 55.95 m IWB Mathematics Pg 434-442 Ex. 17.02 IWB Mathematics Pg 434-442 Ex. 17.02 After what distance travelled is the rocket at a height of 35 metres y =.67x + 20 35 =.67x + 20 15 =.67x x = 22.5 y =.67x + 20

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Practice Assessment – Q2 A school team is investigating selling sausages or discount cards for a fundraiser. Discount card scheme is: - set up costs $10 Projected Earnings E – S + 10 = 0 Sausage scheme – how much would they earn/lose if the y spent $10 on sausages. How much would they have to spend on sausages if they want to earn $50? lose $20 Which is more profitable? Sausages (if over $30 is spent) Steeper gradient. E S E = S – 10 The sausage scheme is shown on the graph E = 2S - 40 spend $45 E = S - 10

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Practice Assessment – Q3 The avg. length of local telephone calls (in min) made in Emilys household over a period time can be modeled by the equation y = -1.5x(x – 16) duration of calls x min a.) Max. occurs at x = y = -1.5(8)(8 – 16) = 96 8 min is the most common duration of call, its occurs 96 times. b.) Longest length of call 16 min 8 c.) Write in VERTEX form of the equation, then shift 2 units to the right and 4 units up. y = -1.5(x – 8) 2 + 96 y = -1.5(x – 10) 2 + 100 subst. y = 0 and solve for x to find length of shortest call (find roots of the equation using GDC) x = 1.84 min or 18.2 min number of calls made y

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Practice Assessment – Q4 (excellence) A telecommunication company undertook an ad campaign to gain clients. The number of new customers signed up daily over 1 month (30 days) is modeled by the parabola shown. Day (x) Day is 7 most successful, 115 new customers (7, 115) Aim is to sign up at least 65 customers per day. Were they successful ? y = k(x – 7) 2 + 115 subst. x = 30 to find the number of new customers on the last day of the campaign y = 66.9, therefore Acme Net was successful in signing up at least 65 customers per day New customers y (18, 104) 104 = k(18 – 7) 2 + 115 k = y = (x – 7) 2 + 115

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finish d (cm) Applications of Parabolas A flea can jump up about 120 times its own body height. A flea jump that goes 10 cm sideways and reaches a height of 15 cm can be modeled by a parabola. The horizontal axis, d, the sideways distance (in cm) and the vertical axis shows the height, h, reached (in cm) What is the equation of the parabola? 0 = k(10 – 5) 2 + 15 -15 = 25k k = -0.6 (10, 0) is on the parabola y = -0.6x 2 + 15 h = k(d – 5) 2 + 15 start h (cm) How tall was the flea? 15 / 120 = 0.125 cm = 1.25 mm

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0 = 0.1(x – 25) 2 250 m Applications of Parabolas The Tower of Terror is a thrill ride. It starts along a horizontal track, reaching a top speed of 160 km/h in 7 seconds, then goes up a curve that can be modeled by a parabola and continues vertically up to the top of a 115 m tower. Then the ride drops in reverse back to the starting point. The equation of the parabolic part of the graph is What are the equations for the vertical and horizontal components of the ride? y = 0.1(x – 25) 2 start Top = 62.5 m 115 m What are the coordinates of D, the y – intercept of the parabola? x = 0 y = 0.1(0 – 25) 2 D What are the coordinates of B, the x – intercept of the parabola? y = 0 x = 25 (0, 62.5) (25, 0) B x = 0, y = 0

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250 m Applications of Parabolas The Tower of Terror is a thrill ride. It starts along a horizontal track, reaching a top speed of 160 km/h in 7 seconds, then goes up a curve that can be modeled by a parabola and continues vertically up to the top of a 115 m tower. Then the ride drops in reverse back to the starting point. Horizontal = ______________ Vertical = _________________ Curve = ______________ How long is the track approximately? y = 0.1(x – 25) 2 start Top = 67.3 m 115 m 250 – 25 = 225 m (25) 2 + (62.5) 2 D What is the total distance you would travel while on this ride? Length 344 m l 365 m x 690 m (0, 62.5) (25, 0) B 115 – 62.5 = 52.5 m

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Applications of Parabolas A rugby player places the ball on the ground 46 metres from the goal posts. He steps back, runs towards the ball and kicks it towards the goal posts. Will he score a goal? The ball must pass over a bar 3 metres above the ground between the goal posts. The balls path through the air can be modeled with a parabola. -36/24 2 = k 0 = k(0 – 24) 2 + 36 y = -0.0625(x – 24) 2 + 36 y = k(x – 24) 2 + 36 Subst. x = 46 to see if ball will clear the goal (0,0) is on the parabola = 5.75 m > 3.00 m yes it will clear 24 m 46 m Maximum height 36 m Goal 3 m high y = -0.0625(46 – 24) 2 + 36 Alternative equation y = -0.0625(x – 24)(x + 24) Subst. point (0, 36)

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