• • • 7.1 R eview of Graphs and Slopes of Lines Standard form of a linear equation:

Ax



By



C

The graph of any linear equation in two variables is a straight line. Note: Two points determine a line.

Graphing a linear equation: 1. Plot 3 or more points (the third point is used as a check of your calculation) 2. Connect the points with a straight line.

7.1 R eview of Graphs and Slopes of Lines • Finding the x-intercept (where the line crosses the x-axis): let y=0 and solve for x • Finding the y-intercept (where the line crosses the y-axis): let x=0 and solve for y Note: the intercepts may be used to graph the line.

7.1 R eview of Graphs and Slopes of Lines • If y = k, then the graph is a horizontal line (slope = 0): • If x = k, then the graph is a vertical line (slope = undefined):

7.1 R eview of Graphs and Slopes of Lines • Slope of a line through points (x 1 , y 1 ) and (x 2 , y 2 ) is:

m

 change in change in x y  (

y

2 (

x

2  

y

1 )

x

1 ) 

rise run

• Positive slope – rises from left to right.

Negative slope – falls from left to right

7.1 R eview of Graphs and Slopes of Lines • Using the slope and a point to graph lines: Graph the line with slope passing through the 5 point (0, 0)

m

 3 5 

rise run

Go over 5 (run) and up 3 (rise) to get point (5, 3) and draw a line through both points.

7.1 R eview of Graphs and Slopes of Lines • • Finding the slope of a line from its equation: 1. Solve the equation for y 2. The slope is given by the coefficient of x Parallel and perpendicular lines: 1. Parallel lines have the same slope 2. Perpendicular lines have slopes that are negative reciprocals of each other

• 7.1 R eview of Graphs and Slopes of Lines 1.

Example: Decide whether the

x

 2

y

lines are parallel, perpendicular, or neither: solving for y

x

 2

y

 7  2

y

  2

x



x

in first equation: 

y

  1 2

x

 7 2 

m

   7 

y

1 2  7  3 2.

3.

solving for y in second equation:  2

x



y



m

 2  3 

y

 2

x

 3 The slopes are negative reciprocals of each other so the lines are perpendicular

7.2 Review of Equations of Lines • Standard form:

Ax

• Slope-intercept form: 

y By

 

mx C



b

(where m = slope and b = y-intercept) • Point-slope form: The line with slope m going through point (x 1 , y 1 ) has the equation:

y



y

1 

m

(

x



x

1 )

7.2 Review of Equations of Lines • Example: Find the equation in slope-intercept form of a line passing through the point (-4,5) and perpendicular to the line 2x + 3y = 6 1. solve for y to get slope of line 2

x

 3

y

 6  3

y

  2

x

 6

y

  3 2

x

 2 

m

  3 2 2. take the negative reciprocal to get the 

m

  3 2 slope

7.2 Review of Equations of Lines • 3.

4.

Example (continued):

m

  3 2 Use the point-slope form with this slope and the point (-4,5)

y

 5 

y

 5  3 2 3 2  

x x

  ( 4   4 )   3 2

x

 6 Add 5 to both sides to get in slope intercept form:

y

 3 2

x

 11

7.3 Functions Relations

• Relation: Set of ordered pairs: Example: R = {(1, 2), (3, 4), (5, 1)} • Domain: Set of all possible x-values • Range: Set of all possible y-values • What is the domain of the relation R?

Domain: x-values (input)

7.3 Functions Relations

Range: y-values (output) Example: Demand for a product depends on its price.

Question: If a price could produce more than one demand would the relation be useful?

7.3 Functions

-

Determining Whether a Relation or Graph is a Function • A relation is a function if: for each x-value there is exactly one y-value – Function: {(1, 1), (3, 9), (5, 25)} – Not a function: {(1, 1), (1, 2), (1, 3)} • Vertical Line Test – if any vertical line intersects the graph in more than one point, then the graph does not represent a function

7.3 Functions

• Function notation: y = f(x) – read “y equals f of x” note: this is not “f times x” • Linear function: f(x) = mx + b Example: f(x) = 5x + 3 • What is f(2)?

7.3 Functions Graph of a Function

• Graph of

f

(

x

) 

x

• Does this pass the vertical line test?

What is the domain and the range?

7.3 Functions - Graph of a Parabola

f

(

x

) 

x

2 Vertex

7.4 Variation

• Types of variation: 1. y varies directly as x:

y



kx

2. y varies directly as the n th power of x:

y



kx n

3. y varies inversely as x:

y



k x

4. y varies inversely as the n th power of x:

y



k x n

7.4 Variation

• Solving a variation problem: 1. Write the variation equation.

2. Substitute the initial values and solve for k.

3. Rewrite the variation equation with the value of k from step 2.

4. Solve the problem using this equation.

• 1.

2.

3.

4.

7.4 Variation

Example: If t varies inversely as s and t = 3 when s = 5, find s when t = 5 Give the equation:

t



k s

Solve for k: 3 

k

5 

k

 15 Plug in k = 15:

t

 15

s

When t = 5: 5  15

s

 5

s

 15 

s

 3

9.2 Review – Things to Remember

• Multiplying/dividing by a negative number reverses the sign of the inequality • The inequality y > x is the same as x < y • Interval Notation: – Use a square bracket “[“ when the endpoint is included – Use a round parenthesis “(“ when the endpoint is not included – Use round parenthesis for infinity (  )

9.2 Review - Compound Inequalities and Interval Notation Solve each inequality for x: 3

x

 1 3

x x

   10 and 2 9 3 and and 2

x x x

   2 2 1  4 Take the intersection: 1  (why does the order change?) Express in interval notation:

x

 3 1 3

9.2 Review - Compound Inequalities and Interval Notation Solve each inequality for x: 2

x x

  3 3

x

  

x

or 0 or 3 or

x x

3

x

   3  1  1 Take the union:

x

  1 Express in interval notation [  1 ,  ) -1

•

9.2 Review - Absolute Value Equations

Solving equations of the form:

ax



b



k

3

x

 4  1 3

x

 4  1 or 3

x

 4   1 3

x

  3 or 3

x

  5

x

  1 or

x

  5 3

9.2 Absolute Value Inequalities

compound inequality (intersection): 

k



ax



b



k

compound inequality (union):

ax



b



k

or

ax



b

 

k

9.2 A Picture of What is Happening

y • Graphs of

f

(

x

) 

ax



b

f(x) = k and f(x) = k ( 

b a

, 0 ) x The part below the line f(x) = k is where

ax



b



k

The part above the line f(x) = k is where

ax



b



k

9.2 Absolute Value Inequalities - Form 1 • 1.

2.

3.

4.

Solving equations of the form: Setup the compound inequality 3

x

 4  1  1  3

x

 4

ax

  1

b

Subtract 4 all the way across  5  3

x

  3 Divide by 3 Put into interval notation 3  5    5 3 ,

x

  1   1 

k

• 1.

2.

3.

4.

9.2 Absolute Value Inequalities - Form 2 Solving equations of the form: 3

x

 4  1 Setup the compound 3

x

 4  inequality

ax

 1 or 3

x b

  4 Subtract 4 all the way across 3

x

  3 or 3

x

  5

k

 Divide by 3

x

Put into interval notation  1     1 or x  ,  5 ) 3    5 3 (  1 ,   What part of the real line is missing?

•

9.2 Absolute Value Inequality that involves rewriting

Example:

x

 2  3   1 Add 3 to both sides (why?):

x

 2  2 Set up compound equation:  2 

x

 2  2 Add 2 all the way across: 0 

x

 4 Put into interval notation

9.2 Absolute Value Inequalities

• Special case 1 when k < 0: 5

x

 3   4 Since absolute value expressions can never be negative, there is no solution to this inequality. In set notation: 

• 9.2 Absolute Value Inequalities Special case 2 when k = 0: 5

x

 3  0 Since absolute value expressions can never be negative, there is one solution for this: 5

x

 3  0 5

x

 3  0 5

x

  3 In set notation:   5

x

  3 5 What if the inequality were “<“?

9.2 Absolute Value Inequalities • Special case 3: 5

x

 3   1 Since absolute value expressions are always greater than or equal to zero, the solution set is all real numbers. In interval notation:    ,  

9.2 A Picture of What Happens When k is Negative

y • Graphs of

f

(

x

) 

ax



b

and f(x) = k ( 

b a

, 0 ) f(x) = k

f

(

x

) 

ax



b

never gets below the line f(x) = k so there is no solution to

ax



b



k

x

9.2 Relative Error

• Absolute value is used to find the relative error of a measurement. If x t represents the expected value of a measurement and x represents the actual measurement, then relative error in

x



x t



x t x

9.2 Example of Relative Error

• A machine filling quart milk cartons is set for a relative error no greater than .05. In this example, x t = 32 oz. so: 32 

x

 .

05 32 Solving this inequality for x gives a range of values for carton size within the relative error specification.

1.

2.

3.

4.

5.

6.

9.2 Solution to the Example

Simplify: 32 

x

32  1 

x

32  .

05 Change into a  .

05  1  compound inequality Subtract 1  1 .

05  

x

32

x

32   .

05  .

95 Multiply by –32 33 .

6 

x

 30 .

4 Reverse the inequality Put into interval notation 30 .

4  

x

 30 .

4 , 33 .

6 33 .

6 