IA

Functions, Equations, and GraphsChapter 2

In this chapter, you will learn:

• What a function is.• Review domain and range.• Linear equations.• Slope. • Slope intercept form y = mx+b.• Point-slope form y – y1 = m(x – x1).• Linear regression.

What is a function?

A function is a special type of relation in which each type of domain (x values) is paired of with exactly one range value (y value).

FUNCTION FUNCTION

NOT A FUNCTION

FUNCTIONFUNCTION

NOT A FUNCTION

NOT A FUNCTION

Relations and Functions

-3

0

2

1

2

4

Suppose we have the relation { (-3,1) , (0,2) , (2,4) }

FUNCTIONONE – TO – ONE

DOMAIN x - values

RANGE y - values

Relations and FunctionsSuppose we have the relation { (-1,5) , (1,3) , (4,5) }

-1

1

4

5

3

5

FUNCTIONNOT ONE – TO – ONE

Relations and FunctionsSuppose we have the relation { (5,6) , (-3,0) , (1,1) , (-3,6) }

5

-3

1

0

1

6

NOT A FUNCTION

Domain and Range

The set of all inputs, or x-values of a function.It is all the x – values that are allowed to be used.

The set of all outputs, or y-values of a function.It is all the y – values that are represented.

Example 1

• Domain = ________________• Range = _________________

All x – values or (-∞ , ∞)

Just 4 or {4}

Example 2

• Domain = ________________• Range = _________________All y – values or (-∞ , ∞)

Just -5 or {-5}

Example 3

• Domain = ________________• Range = _________________

All x – values or (-∞ , ∞)

From -6 on up or [-6 , ∞)

Example 4

• Domain = ________________• Range = _________________All y – values or (-∞ , ∞)

From -6 on up or [-6 , ∞)

Example 5

• Domain = ________________• Range = _________________All y – values or (-∞ , ∞)

All x – values or (-∞ , ∞)

Function Notation

• Function notation, f(x) , is called “f of x” or “a function of x”.

• It is not f times x .

• Example: if y = x+2 then we say f(x) = x+2.• If y = 5 when x = 3, then we say f(3) = 5

What is function notation?

Example 1f(x) = 3x + 1

f( 13) = ____________________

f( 5) = ____________________

f( -11) = ____________________

3 (5) + 1 = 16

3 (13) + 1 = 40

3 (-11) + 1 = -32

Example 2f(x) = x² + 3x - 5

f( 0) = ____________________

f( 5) = ____________________

f( 4) = ____________________

5² + 3 (5) – 5 = 35

0² + 3 (0) – 5 = -5

4² + 3 (4) – 5 = 23

SLOPE

RISE

RUN

SLOPE

Slope Formula

12

12

xx

yy

run

risem

Given points (X1,Y1) and (X2,Y2)

12

12

xx

yy

Is the same as ?

21

21

xx

yy

Example( 4 , 0 ) and ( 7 , 6 )(X1,Y1) (X2,Y2)

12

12

xx

yym

47

06

2

Example( -6 , 5 ) and ( 2 , 4 )

(X1,Y1) (X2,Y2)

12

12

xx

yym

62

54

8

1

Example( 5 , 2 ) and ( -3 , 2 )(X1,Y1) (X2,Y2)

12

12

xx

yym

53

22

8

0

0

Example( 2 , 7 ) and ( 2 , -3 )

(X1,Y1) (X2,Y2)

12

12

xx

yym

22

73

0

10 undefined

Recap of slope

Linear Forms of Linear EquationsStandard Form (AX + BY = C)

1) A and B cannot be fractions.2) A cannot be negative

y = 3x – 5 y – 3x = – 5

– 3x + y = – 5

3x – y = 5

To find y-intercept, set x = 0Y-intercept (0,-5)

FINDING INTERCEPTS

To find x-intercept, set y = 0X-intercept (5/3,0)

Linear Forms of Linear EquationsSlope Intercept Form (y = mx + b)

1) Y is isolated on one side.2) Y is positive

x – 2y = 6 –2y = – x + 6

y = ½ x + 6/-2

y = ½ x – 3

To find y-intercept, set x = 0Y-intercept (0,-3)

FINDING INTERCEPTS

To find x-intercept, set y = 0X-intercept (6,0)

Find a line|| to 2x + 4y = -8

and passes thru (8,3)

• First find slope• 2x + 4y = -8• 4y = -2x – 8• y = - ½ x – 2

• Slope

• Use y = mx + b• y = - 1/2x + b• Plug in point• 3 = - ½ (8) + b• 3 = -4 + b• 7 = b

FINAL EQUATION y = mx + by = - ½ x + 7

Slope intercept

Find a linePerpendicular to 3x – 2y = 10

and passes thru (-6,2)

• First find slope• 3x – 2y = 10 • - 2y = -3x + 10• y = 3/2 x – 5

• Slope

• Use y = mx + b• y = -2/3 x + b• Plug in point• 2 = -2/3 (-6) + b• 2 = 4+ b• -2 = b

FINAL EQUATION y = mx + by = -2/3x - 2

Slope intercept

Find a lineThat has slope = 2

and passes thru (-4,7)

• Use y = mx + b• y = 2x + b• Plug in point• 7 = 2 (-4) + b• 7 = -8 + b• 15 = b

FINAL EQUATION y = mx + by = 2 x + 15

Slope intercept

Find a lineThat has x-intercept = (5,0)

and y-intercept (0,-3)

• Use y = mx + b• y = 3/5 x + b• Plug in point

either one !• 0 = 3/5 (5) + b• 0 = 3 + b• -3 = b

FINAL EQUATION y = mx + by = 3/5 x - 3

Slope intercept• First find slope

• m

• m = 3/5

50

03

Graph x = 3

Graph y = -3

Graph Y = -3/4 x + 2

Y = mx + b m = slope b = y-intercept

1) Plot the y-intercept first 2) Starting from the y-intercept,

go up if + or down if – Then go right. 3) Plot point and draw your line.

m = -3 / 4 b = 2

Graph 5x + 6y < 30

• 5x + 6y < 30• 6y< -5x + 30• Y < -5/6 x +5

Graph 2x + 4y ≥ 16 • 2x + 4y ≥ 16• 4y ≥ -2x + 16• Y ≥ -2/4 x + 4• Y ≥ -1/2 x + 4

Graph 5x - y < 8• 5x – y < 8• – y < -5x + 8• y > 5x – 8

Graph y = |x|

X Y

0 0

1 1

3 3

-3 3

-5 5

WHERE IS THE VERTEX? (0,0)

Graph y = |x - 3|

X Y

0 3

1 2

3 0

4 1

5 2

6 3

WHERE IS THE VERTEX? (3,0)

Graph y = |x + 3|

X Y

0 3

1 4

-3 0

-5 2

2 5

-8 5

WHERE IS THE VERTEX? (-3,0)

Graph y = |x + 3| – 2

X Y

0 1

1 2

2 3

-3 -2

-5 0

-8 3

WHERE IS THE VERTEX? (-3,-2)

Is there a formula for graphing absolute value equations???

• Y = |x + 2| – 3• Y = |x + 5| + 8• Y = |x – 8| – 6 • Y = |x – 7| + 4• Y = |x – 4| – 5• Y = |2x – 8| + 2• Y = |3x + 6| – 3• Y = |mx + b| + c

(-2 , -3)(-5 , 8)(8 , -6)(7 , 4)(4 , -5)(4 , 2)(-2 , -3)(- b/m , c)

Correlationspositive

As x increasesThen y increases

POSITIVE SLOPE

“Trend line” or “regression line”

Outlier

Correlationsnegative

As x increasesThen y decreases

negativeSLOPE

Correlationsnone

No real trend line