Revision: 1.Solving Equations

1.3 ALGEBRA Relationships between tables,

equations or Graphs4 Credits (External)

AS 91028

Gamma Mathematics - Ex 12,13,14 and 15 Page 254 Work book – Chapter 3 Page73

Revision: 1.Solving Equations

To solve equations you must:• Get the unknown to one side of the

equal sign by itself• Do the same operation to both sides

of the equals sign• Start each step on a new line with all

the equals signs lined up

Examples: Solve

x + 5 = 7

x = 2

= -3

542x =

2x

x = 14 x = -6

x = 7 - 5

5x = 42

x - 7 = 7

x = 7 + 7 x = -3 x 2

In an equation with more than one operation, the opposite operations must be applied in reverse order

Solve

6x + 5 = 17

Example:

6x = 12

5x

= 12

23x

= 9 + 3

x = 60

6x = 17 - 5

x = 12 x 5

= 7

x + 3 = 7 x 2

- 3 = 9

x = 2x + 3 = 14

x = 14 - 3x = 11

5x

5x

2. Solving InequationsLinear inequations have an inequality symbol instead of an equals sign:

< less than < less than or equal to

> greater than > greater than or equal to

To solve inequations follow the same procedure as solving equations.

When multiplying or dividing by a negative number swap the inequality sign around.

Examples: Solve

3x + 5 > -7

3x > -123x > -7 - 5

-5x - 2 < -22

x > -4

-5x < -22 + 2

-5x < -20

x > 4

Dividing by a negative – swap sign!!

Note: always check your answer by putting putting in a number to check that the equation is true!!

More examples:

-2x - 4 > 10

x < -7-2x > 14

-7 – 3x < -1 -3x < 6

x > -2

3. Substitution

Substitution means replacing a symbol with a value. Remember to follow the rules of BEDMAS.

Examples: Calculate the value of these expressions 7x – 1 when x = 27 2 – 1= 14 – 1

= 13

when f = 2 and g = 6

= 4

5x2 - 3x + 2 when x = -35x(-3)2 - 3 -3 + 2

= 56= 45 - -9 + 2

Exercise 12.01

Note 1: Linear PatternsLinear Patterns are sequences of numbers where the difference between successive terms is always the same.

The general rule for a LINEAR pattern is always in the form:

Term y = dx + c

x is the position of the term in the sequenced is the differencec is a constant

Example 1:Find the rule that generates the sequence

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 + CC = 4 - 3C = 1

Rule to find each term is y = 3x + 1

e.g. 2 Identify if the following relations are linear.

x Y-2 7-1 40 11 -22 -53 -8

Diff. x Y-2 2-1 40 81 162 323 64

Diff.

Common difference = -3Therefore, Linear

No Common difference Therefore, Not Linear

-3

-3-3-3-3

2

481632

e.g. Find the differences, and if the result is linear find the rule.

Term # Sequence

1 162 133 104 75 46 1

Diff.

Common difference = -3Therefore, Linear

-3-3-3-3

Each Term = −3 x Term # + C16 = −3 x 1 + C16 = −3 + C19 = C

Each Term = −3x + 19Rule: y = -3x + 19

(X)

-3

Check that your formula works !

Find the differences, and if the result is linear find the rule.

Term # Sequence

1 -82 -63 -44 -25 06 2

Diff.

Common difference = Therefore,

2222

Each Term = 2 x Term # + C -8 = 2 x 1 + C

-8 = 2 + C-10 = C

Each Term = 2x − 10Rule: y = 2x - 10

(X)

2


NOW YOU TRY THIS ONE -8, -6, -4, -2, 0, 2……

+2linear

Find the differences, and if the result is linear find the rule.

Term # Sequence

1 62 113 164 215 266 31

Diff.

Common difference = Therefore,

5555

Each Term = 5 x Term # + C 6 = 5 x 1 + C6 = 5 + C1 = C

Each Term = 5x + 1Rule: y = 5x + 1

(X)

5


NOW YOU TRY THIS ONE 6, 11, 16, 21, 26, 31……

5linear

HMWK pg 74Ex A

IWB Mathematics Pg 269-273 Ex. 12.02 # 3 -12

Using Rules for Linear Sequences

Examples:The rule for a linear sequence is t = 4n -2Find the 7th term of the sequence

t = 4 x 7 – 2=26

Which term has a value of 74?74 = 4n -276 = 4n n = 19

If the rule for a linear sequence is known, then the values of terms or the term number can be found algebraically.

What is the first term in the sequence that has a value over 151?

4n – 2 > 151 4n > 153 n > 38.25

39th term is the first term greater than 151

IWB Mathematics Pg 269-273 Ex. 12.02

HMWK pg 74-82Ex B

y = 2x – 3

xy = -13

y = x(x + 2)

y = 2

Linear y1 = 2x1 - 3

Not Linear x = -13y-1

Not Linear y1 = x2 + 2x1

Linear y1 = 2

Starter Identify if the following relations are linear.

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 x2, x3, x-1 etc)There is a common difference between ordered pairs.

Plot points from the table on an axis.

(x, y)

Across the x-axis Up/Down the y-axis

x -2 -1 0 1 2 3

yCo-

ordinate (-2, ) (-1, ) (0, ) (1, ) (2, ) (3, )

y

x1 2 3 4-1-2-3

12345

-1-2-3-4-5

y = 2x - 1

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 = -2 y = 2 -2 – 1 = -5

When x = -1 y = 2 -1 – 1 = -3

-5 -3 -1 1 3 5

Note 2: Graphs of linear functions

HMWK pg 78Ex C and D

IWB Mathematics Pg 295 Ex. 13.01

Note 3: 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 = )x(run

)y(rise

Horizontal Change run

Vertical Change rise

Note 3: 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

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 3: Gradients

e.g. Draw lines with the following gradients. a = b = 3 2

3

Note 3: Gradients

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.

31

93

=

Note 3: Gradients

e.g. Find the gradients of the following lines.

a b c d e f

25

42

12

11

21

b=

undefined

GAMMA Pg 298 Ex. 13.02

21

=

Note 3: Gradients

HMWK pg 83Ex E

Note 4: Straight Line Graphs

The general formula for a straight line is

y = mx + c

There are a couple of straight lines that do not fit this general formula:

y = number gives a horizontal linex = number gives a vertical line

gradient y-int

e.g. y = 2 x = 4

y

x1 2 3 4-1-2-3-4

1234

-1

y

x1 2 3 4 5

1234

-1-2-3-4


To draw straight line graphs using y = mx + c

• Mark the y-intercept (given by c). If there is no c, the y-intercept is 0.

• From the y-intercept, count the gradient (given by m) Put a point and then count the gradient again. If there is no m, it is always 1.

• Rule the straight line between the three points.

• Put arrows on both ends of the line and label it.


e.g. Draw the following straight lines.

a) y =

b) y = 4x – 2

c) y = x + 1

d) y = -2x

221

x

x

y 221

xy = y = -2x y = 4x - 2

y = x + 1


GAMMA Pg 306 Ex. 13.03

HMWK pg 86Ex F

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 = 0y = -4 (0) + 12y = 12 therefore the y intercept is (0, 12)Find x-intercept, set y = 00 = -4x + 124x= 12x = 3 therefore the x intercept is (3, 0)

Find x-int, Set y = 02x – 0 = 6 2x = 6 x = 3 therefore the x-intercept is (3, 0)Find y-int, Set x = 02(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

y

x2 4 6 8

2

4

6

8

-2

-4

-6

-8

x-intercept is (3, 0)

y-intercept is (0, -6)

The cover up method – a quick way of graphing a straight line in the form ax + by = c

-x + 2y = 8Is to cover up the x term to get the y-intercept

-x + 2y = 8And then cover the y term to get the x-intercept

-x + 2y = 8Then plot the 2 points.

i.e. y = 4 (0,4)

i.e. x = -8 (-8,0)

x

y

(0,4)

(-8,0)

-x + 2y = 8

IWB MathematicsEX 13.04 pg 310

HMWK G, H

Starter – Graph to Equation

y = 2

x + 3

y = -1/2 x + 0

y = 1/3x - 5

y = -4x + 4

2

1

3

4

Writing Equations for GraphsBy looking at a graph we can write the equation for a line by finding the:

y-intercept (c)Gradient (m)

The equation for the graph is found by replacing m and c into the general format for a line: y = mx + c.

For a vertical line: x = c For a horizontal line: y = c

Where c is a constant

Equation ac = 2m = -¾y = -¾x + 2

Equation bx = 5

Equation cc = -6m = 6/4 = 3/2 y = 3/2x - 6

Example: Write the equations for the lines

StarterDraw the following straight lines.

3x + 6y = 6

x

y

3x + 6y = 6

y = 03x = 6x = 2Int: (2, 0)

y = 02x = 4x = 2Int: (2, 0)

x = 06y = 6y = 1Int (0, 1)

x = 0-y = 4y = -4Int: (0, -4)

2x - y = 4

2x – y = 4

Note 5: Applications of the Straight Line GraphMany circumstances in nature and life can be graphed using straight line relationships.

The y-intercept represents the value of y initially (when x=0)Example: A fixed administration fee

The gradient represents the rate at which y changes compared with x.Example: An hourly hire charge

Note 5: Applications of the Straight Line Graph

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 20’s).

y = 10x + 50.

x

yCost $

Number of Pizzas sold8642 10 1412

120

100

80

60

40

20

140

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

50If no pizzas are sold, it still costs them $50

10 Represents the fact that each pizza they make, costs $10

y

x 2 4 6 8 10

20

40

60

80

100

120

140

160

Number of Pizzas sold

Cost

$

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)

Dist

ance

(km

)

The gradient of the graph is also a measure of the rate of change of the dependent variable

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

2000

2002

2004

2006

2008

2010

Before 2004

After 2004 = 400 – 330 2010 – 2004 = 70 6 = $11 667/yr

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

10 20 30

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

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

C100 km/hr20 km/hr-65 km/hr

100t - 400

4.68 hrs = 4hrs 41mins

IWB MathematicsEx 13.06 Pg 316 - 321

Work bookEx I

StarterDraw a distance-time graph to describe the fable of the tortoise and the hare. (assume 3 km race)

Time taken (hours)

Dist

ance

from

star

t (km

)

1

32

2

3

1 4

Note 7: Simultaneous EquationsThe point (x,y) where two linear equations cross, gives a unique solution for x and y that satisfies both equations.

105

5

10

15

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

Note 7: Simultaneous EquationsAn 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 MathematicsEx 13.05 Pg 313-314

2d + 6c = 12 (2)4d + 4c = 16 (1)

Dogs need 3 cans and cats need 1 can of food.

Starter

(0,2)

4y = 8y = 2

Subst. y = 2 into either equation2x + 3(2) = 6

2x = 0x = 0

Note 8: Quadratic Patterns

Quadratic patterns have a constant second order of differences (differences of differences).

Term #

Sequence

1 52 113 214 355 536 75

Diff.Diff.

610

141822

444

4

The value of the 2nd order of differences determines our Rule

2nd

1 means a 0.5n2

2 n2

3 1.5n2

4 2n2

In general 2nd Diff n2

2

Now that we know what the constant for n2

we can solve for the rest of the equation by substituting the term # into ____________

Term # (n)

Sequence 2n2 Adjustment

1 52 113 214 355 536 75

2n2

2818325072

333333

Rule: 2n2 + 3

NOW YOU TRY 4, 6, 10, 16, 24, 34….

Quadratic patterns have a constant second order of differences (differences of differences).

Term #

Sequence

1 42 63 104 165 246 34

Diff.Diff.

24

6810

222

2

The value of the 2nd order of differences determines our Rule

2nd

1 means a 0.5n2

2 n2

3 1.5n2

4 2n2

In general 2nd Diff n2

2

Now that we know what the constant is for n2

we can solve for the rest of the equation by substituting the term # into _______________

Term # (n)

Sequence n2 Adjustment

1 42 63 104 165 246 34

n2

149162536

3210-1-2

Rule: n2 – n + 4

HMWK pg 94-100

IWB Mathematics Pg 277-281Ex. 12.03

ParabolasFound in nature, motion, structure & art

Note 9: Quadratic Functions

Quadratic functions are algebraic expressions where the highest power of x is x2

e.g. y = x2

y = x2 + 3x – 5y = (x-2)(x+2)y = (x+2)2

the graph of these functions are called parabolas.

y = x2

vertex (0,0)

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

x y =x2

-3

-2

-1

0

1

2

3

9

4

1

0

1

4

9

y = x2

Transformations of the parabola

y = x2 + 7

y = x2 + 4

y = x2

y = x2 − 3

By adding/ subtracting a constant, the parabola shifts up /down.

1.) Vertical Translation y = x2 ± c

y = x2

By adding/ subtracting a constant to the x variable, the parabola shifts left /right.

2.) Horizontal Translation y = (x±k)2

y = (x−3)2 y = (x+5)2 y = (x+2)2

y = x2

3.) Combined – Horizontal & Vertical Translation y = (x+k)2 + c

y = (x−4)2+2

y = (x+3)2−4


To get the y-int:Let x = 0

WorkbookPg 104 Ex. N

StarterWrite the equations for each of the parabolas drawn below

y = (x+6)2 - 3y = (x-4)2 - 5

y = (x-5)2 + 4

Note 10: Parabolas – Vertical Scale Factors

If the coefficient of x2 is negative, then the parabola is inverted (upside down)

e.g. y = −x2 + 2y = −(x + 3)2

y = −(x + 3)(x − 2) y = (x + 3)(2 − x)

Different forms of the same equation

y = -x2

The parabola y = kx2 transforms the basic parabola:

If k is greater than 1, the parabola is thinner and steeper

If k is between -1 and 1, the parabola will be wider and more shallow

If k is negative the graph is upside down

Parabolas with a Change of Scale

y = x2

y = 2x2

Examples:

2x41y

y = -3x2

y = 2(x – 1)2 – 2

Vertex = y – intercept, when x =

(1, -2)0

Across 1, up 1 x2 = 2Across 2, up 4 x2 = 8Across 3, up 9 x2 = 18

y = 2(0 – 1)2 – 2 = 0 (0, 0)

y = −x2

y = −2x2

y = −3x2

y = x2

y = −x2

y = 1/2x2

y = x2

y = 1/4 x2

y = 1/10 x2


HMWK pg 104-107

Starter: Graph the following parabola

y = 2(x – 1)2 – 2

* Opens upwardVertex = y – intercept =

(1, -2)0

Over 1, up 1 x2 = 2

Over 2, up 4 x2 = 8

Over 3, up 9 x2 = 18

Step by step for graphing factorised parabolas

1.) Find y-int by setting x = 02.) Find x-int by setting y = 03.) 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.

Note 11: Parabolas (factorised form)

Example

Plot the parabola for: y = (x - 4)(x + 2)

Y-int – set x = 0 X-int – set y = 0

y = (0 - 4)(0 + 2)

y = -4 x 2y = -8

0 = (x - 4)(x + 2)x - 4 = 0 or x + 2 = 0 x = 4 x = -2

Graph of y = (x – 4)(x + 2)

x-int

y-int

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

Axes of symmetry

Graph the following parabolas

a.) y = (x – 3)(x +4) b.) y = x(x - 6)

y-int (x = 0) y-int (x = 0)

y = -3 x 4 y = 0 = - 12

x-int (y = 0) x-int (y = 0)

0 = (x – 3)(x + 4) 0 = x (x - 6)x – 3 = 0 x + 4 = 0 x = 0 x - 6 = 0 x = 3 x = -4 x = 6

y = (x – 3)(x +4)

y = x(x - 6)

(3, -9)

(-0.5, -12.25)


HMWK pg 108-109

Equations can be found by examining the features of the graph and using the

following methods:

Transformation MethodFactored Form Method

Note 12: Writing Equations of a Parabola

Transformation MethodFind the vertex of the parabola (c, d), where c is the shift on the x-axis and d is the shift on the y-axis.The equation is:

y = (x – c)2 + dIf the parabola is upside down the equation is:

y = - (x – c)2 + d

Factored Form MethodFind the x-intercepts of the parabola (the points c and d).The equation is:

y = (x – c)(x – d)If the parabola is upside down the equation is:

y = - (x – c)(x – d)

y = (x – 1(x – 3)this format shows the x-intercepts, 1 and 3

y = (x – 2)2 – 1This format gives the turning point (2, – 1)

Example: Find the equation of the line in both formats

y =- (x+7)2 + 2

2

1

y = x2 - 5

y = (x+5)(x-1)

1

2

y = −(x-2)(x-4)


HMWK pg 112-113Ex Q

Check your intercepts are correct once you have composed an equation

Equations can be found by:

Counting SquaresSolving algebraic equations

Note 13: Writing Equations of a Parabolawith a Scale Factor

Example: Write Equations for1

2

Graph 1 - Factored Formy = k(x + 3)(x + 9)To find k:* Count squares across 3, down 3 – should be 9 upside down - negative k = -1/3

* Substitution using the vertex (-6,3):

3 = k(-6 + 3)(-6 + 9)3 = -9kk = -1/3

Graph 2 - Transformed Formy = k(x - 4)2 - 5To find k:* Count squares across 1, up 2 – should be 1 k = 2* Substitution using another point (2,3):

3 = k(2 - 4)2 -53 = 4k - 5k = 2

Graph to Equation– Vertical scale factor

y =-1/3 (x+9)(x+3)

1

2

y = 2(x-4)2 - 5


y =-1/3 (x+6)2 + 3

HMWK pg 112-113


Starter

C EA

D BF

Starter

FA

CE

N K J

H

DP

Q

B

Applications to ParabolasA 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

Applications to ParabolasA 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 pointy = -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 okYes, there is sufficient room for 12 courier vans

Starter

B F ED

C

J I H G

A

Graph to Equation in form y=(x+k)2+c

y =- (x+7)2 + 2

2

1

y = x2 - 5

Graph to Equation in form y=(x+a)(x+b)

y = (x+5)(x-1)

1

2

y = −(x-2)(x-4)


Task: Graph the following parabolas

y = -(x – 5)(x – 1) y = 2(x – 1)2 – 2

* Invertedy – intercept = x – intercepts = Midpoint is x = Vertex = (3, )

-55, 13

4

Starter

=

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

gradient of DE = xy

12

12

xxyy

1552

43

=

=

Note 13: Exponential Relationships

• Commonly known as the ‘growth curve’

• General form is y = ax

• Curve always passes through (0, 1) • x-axis is an asymptote (graph approaches, but never

touches)• this year we only explore when is a natural

number and > 1

1 = a0

base

exponent

a a

Graph by plotting points

x value y = 2x

-3-2-10123

0.1250.250.51248

y = 2x

y=2x+2

the exponential function can be vertically translated up or down by adding or subtracting a constant

HMWK pg 119-120


StarterWrite the rule for the following sequences

4, 8, 12, 16, 20, 24, ……

10, 26, 48, 76, 110, 150, ……

4n

1st Difference

2nd Difference16 22 28 34 40 6 6 6 6 3n2

3n2 3, 12, 27, 48, 75, ……. Difference 7 14 21 28 35

+ 7n

Note 14: Exponential Patterns

Exponential patterns do not have a constant difference between terms

Term #

Sequence

1 32 53 94 175 336 65

Diff.Diff.

24

81632

4÷2 = 28÷4 = 2

16÷8 = 2

32÷16 = 2

The ratio of the 1st order of differences determines our Rule

Ratio of 1st

2 means a base 2 2x

3 …… 3x

4 …… 4x

Now that we know our rule is of the form 2x

we can solve for the rest by substituting the term # into _____

Term # (n)

Sequence 2x Adjustment

1 32 53 94 175 336 65

2x

248163264

111111

Rule: 2x + 1

What will the value of term 9 be?

29 + 1= 513

HMWK pg 116-118IWB Mathematics Pg 286-290 Ex. 12.04

y-intercept = (0,-4.5)

Find an equation for the following parabola in the form y = k (x – c)(x – d)

Applications to ParabolasJacqui 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 Post

Height (m)3

2

1

Supports

1 2-2 -1 0Width (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.

Applications of Parabolas

a.) Parabola is of the form y = kx2 + c

‘At its lowest point it will hang 2.2 metres above the paving.’ Therefore c = 2.2

y = kx2 + 2.2

Parabola is of the form y = kx2 + 2.2

A point that satisfies this equation is (2,3) (-2,3)

3 = k(2)2 + 2.20.8 = 4k

k = 0.2 y = 0.2x2 + 2.2

When x = 1 y = 0.2 (1)2 + 2.2 y = 2.4

(1, 2.4)

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 isxrunyrise

12

12

xxyy

124.22

= − 0.4

22

xy = − 0.4

Multiply both sides by (x-2) y – 2 = - 0.4 (x − 2) y – 2 = - 0.4x + 0.8

HMWK pg 114-115


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 eq’n

4 = k (1 +6)2 - 84 = k (7)2 - 8

12 = 49k

k = 0.245

y = 0.245 (x +6)2 - 8

Find an equation for the following parabola in the form y = k (x – c)2 + dand then find the y-intercept

y = k (x – 3)2 + 13

Subst (4.22,3.57) into eq’n

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 + 13y = -6.34 (9) + 13y = -6.34 (9) + 13y = -44.0

(0,-44)

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 + 2y = -2.2 (x-5)2 + 2y = -2.2 (x+1)2 + 6

y-intercept Set x = 0

y = -2.2 (0+1)2 + 6

y = 3.8

(0, 3.8)


Applications to ParabolasThe 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 12months

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 + 2020 = k(-5)2 k = 0.8

y = 0.8(x-8)2 + 20


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 = -40t2 + 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

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 x2 – 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 x2 – 27 24.5 = 1/3 x2

c.) What is the distance away from the centre when d = -2.5?

73.5 = x2

x = ±8.57 cm (3 sf)

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 x2 – 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 x2 – 27

Revision: 1.Solving Equations

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