relations math 314 time frame slope point slope parameters word problems

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Page 1: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

RelationsRelations

Math 314Math 314

Page 2: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Time FrameTime Frame

• Slope

• Point Slope

• Parameters

• Word Problems

Page 3: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

SubstitutionSubstitution

• Sometimes we look at a relationship as a formula

• Consider 2x + 8y = 16

• We have moved away from a single variable equation to a double variable equation

• It cannot be solved as is!

Page 4: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

SubstitutionSubstitution

• If we know x = 4

• 2x + 8y = 16

• 2(4) + 8y = 16

• 8 + 8y = 16

• 8y = 8

• y = 1

Page 5: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Substitution Substitution • We could say that the point x = 4 and y = 1 or

(4,1) satisfies the relationship. • Ex #2. Given the relationship 5x – 7y = 210, use

proper substitution to find the coordinate (2,y)• (2,y) 5x – 7y = 210• 5(2) – 7y = 210• 10 – 7y = 210• -7y = 200• y = - 28.57• (2, -28.57)

Page 6: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Substitution Substitution

• Ex. #3: Given the relationship 8x + 5y = 80 (x,8)

• (x,8) 8x + 5y = 80• 8x + 5(8) = 80• 8x + 40 = 80• 8x = 40• x = 5 • (5,8)

Page 7: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Substitution Substitution

• Ex: #4 Given the relationship y= 3x2 – 5x – 2

• (-3,y) • (-3,y) y = 3 (-3)2 – 5 (-3) – 2• y = 3 (9) + 15 – 2• y = 40• (-3,40) • Stencil #2 (a-j)

Page 8: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Substitution Substitution

• Given the relationship

Page 9: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Linear RelationsLinear Relations

• We recall…

• Zero constant relation – horizontal

• Direct relation – through origin

• Partial relation – not through origin

• The characteristic here is the concept of a straight line – a never changing start and where it crosses the y axis

Page 10: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example Example

• Line A• Line B

• We say line A has a more of a slant slope or a steeper slope

• (6 compared to 2 is steeper or -6 compared to -2 is steeper).

Page 11: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Variation RelationsVariation Relations

Name of Relation Formula Graph

Direct Relation y = mx

Partial Relation y = mx + b

Zero Variation y = b

Inverse Variation y = m

x

Page 12: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

SlopeSlope

• What makes a slope?

Rise

Run

• We define the slope as the ratio between the rise and the run

• Slope = m = rise

run

Page 13: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Formula for SlopeFormula for Slope

• If we have two points (x1, y1) (x2, y2)

• Slope = m = y1 – y2 = y2 – y1

x1 – x2 x2 – x1

• Remember it is Y over X!

• Maintain order

Page 14: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

1 2 3 4 5 6 7 8-1-2

1

2

3

4

5

-1

-2

-3

-4

-5

x

y

A (x1, y1)

B(x2, y2)

SlopeSlope

• Consider two points

A (5,4), B (2, 1) what is the slope?

Page 15: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Calculating SlopeCalculating Slope

• Slope = m = y1 – y2 = y2 – y1

x1 – x2 x2 – x1

(5, 4) (2, 1) 4 - 1 5 - 2 3 3m = 1

(x1,y1) (x2,y2)

Page 16: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Ex # 2Ex # 2 A = (-4, 2) B=(2, -4)

(x1,y1) (x2,y2) -4 – 2

2 - - 4

- 6

6

m = -1

Page 17: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

1 2 3 4 5-1-2-3-4-5

1

2

3

4

5

-1

-2

-3

-4

-5

x

y (4, 5)

(1, 1)

3

414

15slope

Ex #3Ex #3

(x2,y2)

(x1,y1)

21

21

xx

yyslope

Page 18: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Understanding the SlopeUnderstanding the Slope

• If m or the slope is 2 this means a rise of 2 and a run of 1 (2 can be written as 2 ) 1• If m = - 5, this means a rise of -5 and right

1• If m= -2 this means rise of -2 right 3 3 • Rise can go up or down, run must go right

Page 19: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Consider y = 2x + 3Consider y = 2x + 3

• What is the slope, y intercept, rise & run?

• We can write the slope 2 as a fraction 2

1

• We have a y intercept of 3

• This means rise of 2, run of 1

• Look at previous slide for slope of 4/3

Page 20: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

1 2 3 4 5-1-2-3-4-5

1

2

3

4

5

-1

-2

-3

-4

-5

x

y

Ex#1: y=2x+3Ex#1: y=2x+3

0,3

(1,5)

Question: Draw this line

What is the y intercept?

What is the slope

What does the slope mean?

Where can you plot the y intercept?

Up 2, Right 1

Page 21: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

1 2 3 4 5-1-2-3-4-5

1

2

3

4

5

-1

-2

-3

-4

-5

x

y

(-4, 2) (2,2)

06

0

)4(2

22

slopeIf a line//If a line//xx--

axisaxis

slope = 0slope = 0

ExampleExample

• What do you think the slope will be; calculate it.

Page 22: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

1 2 3 4 5-1-2-3-4-5

1

2

3

4

5

-1

-2

-3

-4

-5

x

y

(2,-3)

(2,2)0

522

32

slope

If a line // If a line // yy--axis:axis:

slope is slope is undefinedundefined

ExampleExample

zero!

Page 23: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

In Search of the EquationIn Search of the Equation

• We have seen that the linear relation or function is defined by two main characteristics or parameters

• A parameter are characteristics or how we describe something

• If we consider humans, a parameter would be gender. (We have males & females). There can be many other parameters (blonde hair, blue eyes, etc.)

Page 24: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

In Search of the Equation NotesIn Search of the Equation Notes

• The parameters we are concerned with are…

• Slope = m = the slope of the line

• y intercept = b = where the line touches or crosses the y axis (It can always be found by replacing x = 0)

• x intercept = where on the graph the line touches or crosses the x axis. (let y = 0)

Page 25: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

In Search of the Equation NotesIn Search of the Equation Notes

• We stated in standard form the equation for all linear functions by y = mx + b. Recall…

• y is the Dependent Variable (DV)• m is the slope• x is the Independent Variable (IV)• b is the y intercept parameter• The key is going to be finding the specific

parameters.

Page 26: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

General FormGeneral Form

• You will also be asked to write in general form

• General Form Ax + By + C = 0

• A must be positive

• Maintain order x, y, number = 0

• No fractions

Page 27: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

General Form PracticeGeneral Form Practice

• Consider y = 6x – 56

• -6x + y + 56 = 0

• 6x – y – 56 = 0

Page 28: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Standard & General FormStandard & General FormExample #1Example #1

• State the equation in standard and general form. • Consider find the equation of the linear function

with slope of m and passing through (x, y).• m = -6 (-2, -3)• (-2, -3) -3 = -6 (-2) + b• -3 = 12 + b• -15 = b• b = -15

Page 29: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example #1 Solution Con’tExample #1 Solution Con’t

• y = -6x – 15 (Standard)

• Now put this in general form

• 6x + y +15 = 0 (General)

Page 30: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Standard & General Form Ex. #2Standard & General Form Ex. #2• m = -2 (5, - 3) 3 • -3 = (-2) (5) + b 3• -3 = -10 + b 3• -9 = -10 + 3b• 1 = 3b• b = 1/3• y = -2 x + 1 (SF)

3 3

• Now General form• Get rid of the

fractions; how?Given y = -2 x + 1

3 3…

Anything times the bottom gives you the top

• 3y = -2x + 1• 2x + 3y – 1 = 0

Page 31: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Standard and General Form Ex #3 Standard and General Form Ex #3

• m = 4

5 (-1, -1)• -1 = 4 x + b

5• -5y = -4x + 5b• 5 (-1) = 4 (-1) + 5b• -5 = -4 + 5b • -1 = 5b • b = -1/5

• y = 4x – 1

5 5• 5 x – 1/5 (standard

form)• 5y = 4x – 1• -4x +5y + 1 = 0 • 4x – 5y – 1 = 0

(general form)

Page 32: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

The Point Slope Method Con’tThe Point Slope Method Con’t

• Consider, find the equation of the linear function with slope 6 and passing through (9 – 2).

• Take a look at what we know based on this question.

• m = 6

• x = 9

• y = -2

Page 33: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Finding the Equation in Standard Finding the Equation in Standard FormForm

• We know y = mx + b• We already know y = 6x + b• What we do not know is the b parameter or the y

intercept• We will substitute the point • (9, -2) - 2 = (6) (9) + b• -2 = 54 + b• -56 = b• b = - 56• y = 6x – 56 (this is Standard Form)• Standard from is always y = mx + b (the + b part can be

negative… ). You must have the y = on the left hand sides and everything else on the right hand side.

Page 34: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

General FormGeneral Form

• In standard form y = 6x – 56

• In general form -6x + y + 56 = 0

• 6x – y – 56 = 0

Page 35: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example #1 8a on StencilExample #1 8a on Stencil

• In the following situations, identify the dependent and independent variables and state the linear relations

• Little Billy rents a car for five days and pays $287.98. Little Sally rents a car for 26 days and pays $1195.39.

• D.V $ Money $

• I.V. # of days

Page 36: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example #1 Soln Con’tExample #1 Soln Con’t

• Try and figure out the equation

• y = mx + b (you want 1 unknown)

• (5, 287.98) (26, 1195.39)

• m = (287.98 – 1195.39)

5 – 26

• m = 43.21

Unknown Unknown

Page 37: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example #1 Soln Con’tExample #1 Soln Con’t

• Solve for b…

• y = mx + b

• (5, 287.98) 287.98 = 43.21 (5) + b

• 287.98 = 216.05 + b

• 71.93 = b

• b = 71.93

• y = 43.21x + 71.93

Page 38: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example #2 8 b on StencilExample #2 8 b on Stencil

• A company charges $62.25 per day plus a fixed cost to rent equipment. Little Billy pays $1264.92 for 19 days.

• I.V. # of days

• D.V. Money

• m = 62.25

Page 39: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Example #2 8a Soln Example #2 8a Soln

• y = mx + b

• (19, 1264.92) 1264.92 = 62.25 (19) + b

• 1264.92 = 1182.75 + b

• 82.17 = b

• b = 82.17

• y = 62.25x + 82.17

Page 40: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Solutions 8 c, d, eSolutions 8 c, d, e

• 8c) IV # of days; DV $

• y = 47.15x + 97.79

• 8d) IV # of days; DV $

• y = 89.97x + 35.22

• 8e) IV # of days DV $

• y= 45.13x + 92.16

Page 41: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Homework HelpHomework Help

• What is the value of x given• 3 = 1 + 1 4 2 x• Eventually, x on the left side, number on the

right side• 3 – 1 = 1 4 2 x• 6x – 4x = 8• -2x = 8• x = -4

• Important step to understand

Page 42: Relations Math 314 Time Frame Slope Point Slope Parameters Word Problems

Homework HelpHomework Help

• What is the opposite of ½ ? • Answer is –½ • If asked what is the opposite of subtracting two

fractions… i.e. ¼ - ½ , find the answer (lowest common denominator and then reverse the sign.

• When told price increases 10% each year… calculate new price after year 1 and then multiply that number by .1 again to calculate price increase for year 2. For example, you have $100 and increases 10%. After year 1 $110 (100 x .1 + 100) & after year two $121 (110 x .1 + 110).