Download 3.2 The Slope of a Line 3.3 The Equation of a Line

(note we are also covering both 3.2 and 3.3 intermixed in these sections)
3.2 The Slope of a Line
The convention is to read a graph left to right.
The slope of a graph is the rise (change in y) divided by the run (change in x) positive-up, negative-down
The formula for slope given two points (x1,y1) and (x2,y2) is m =
y2
x2
y1
, where x1 ≠ x
x1
The Equation of a Line in Slope Intercept Form ( y = mx + b)
Given y-Intercept and slope (If that were only the case! Put the slope in for m and the y-value in b)
Example: What is the equation if the slope was -2 and the y-intercept was (0, 3). y = -2x + 3
What if instead of the y-intercept we had another point (x1, y1) and the slope (Point Slope Form)
y – y1 = m ( x – x1 )
e.g. m = 3 and (2, 4); y – 4 = 3 ( x – 1); simplifies to y = 3x + 1
What if we only had two points (x1,y1) and (x2,y2)
Then we would use the slope formula to determine m, then use the Point Slope Form with either point
e.g. Given (3, 5) and (1, 9); m = (5 – 9)/(3 – 1) = -2; y – 5 = -2 ( x – 3 ); y = -2x – 1
The Standard (or General) Form of a line is Ax + By = C, where A,B,C are integers, A and B both not 0
e.g. 3x – 2y = 6
Note: If you had the equations x = constant (e.g. x = 3), the line would be vertical and the slope undefined
Determine the slope and y-intercept and graph the equation.
20) y = 2x + 3
28) y = 3/5 x – 8
32) y = x – 3
36) 5x + 3y = 6
42) 2x + y = 0
**) y = 7
**) x = 2
Find the slope of the line through the given points.
44) (1,7), (3,1)
48) (-7.9), (2.3)
50) (-3,1). (-3,7)
52) (-5,-1), (-3,-1)
3.3 The Equation of a Line
The primary forms for a line are Slope-Intercept Form (y = mx + b) and Standard Form (Ax + By = C).
To determine the slope of a line from a graph, pick two points and use the slope formula.
Most pairs of straight lines cross at one point. Certain distinct lines don’t cross – parallel lines
Parallel lines have the same slope, but different intercepts.
To determine if two equation are parallel, determine slope. If slopes equal, intercepts not, then parallel
Perpendicular lines ( ┴ ) cross at right angles (90°), Their slopes are flip/flop change signs
e.g. m = 4, then m┴ = - ¼, m = -½ , the m┴ = 2
To determine the equation of a perpendicular line, determine perpendicular slope, use Point Slope Form
Note:
Two horizontal lines or two vertical lines are parallel.
A horizontal line and a vertical line are perpendicular.
Use the slope and the coordinate of the y-intercept to write the equation in Slope Intercept From:
6) m = 3; (0, -1)
10) m = - ¼ ; ( 0, ½)
12) m = 1.6; (0, 4.5)
Write the equation in Slope Intercept Form of the line with the given slope going through the given point.
22) m = 3; (3, 5)
26) m = -1; (0, 0)
30) m = 4/3; (-5, -9)
Write the equations in a) Slope intercept Form and b) Standard Form of the equation going through
34) (1, 3); (-2, -9)
38) (-4, -7); (2, 8)
44) (-5, -2); (6, -4)
Determine if the given set of lines are parallel, perpendicular, or neither
46) y = 2/5 x - 2
y = 2/5 x + 2
48) y = - 2/5 x + 4
y = 5/2 x – 6
50) y = - 1/4 x + 5
y = -4 x
56) y = 3
x=3
**) y = 1
x = -4
Write the equation in Slope Intercept Form that passes through the given point and is parallel to given line
58) (-1, 1); y = -3x + 1
64) (1. 1); 3 x – 4 y = 8
Write the equation in Slope Intercept Form that passes through the given point and is perpendicular to given line
68) (-1, 5); y = 1/2 x – 7
74) (2, 8); x + 2 y = 9
Section 3.4 Graphing Linear Inequalities
A linear inequality is an inequality that can be written
in the form Ax + By > (could also be <, >, <) C, where A, B,
and C are real numbers and A and B are both not zero.
< and > are called strict inequalities and do not include the line (points).
The equation Ax + By = C is the boundary between valid and invalid solutions.
The line is a valid solution for < or > and drawn solid. For < or > it is not valid and drawn as a dashed.
Method to Graph Linear Inequalities
a. Determine three points on the graph of Ax + By = C.
b. If inequality is < or >, draw a dashed line through the points. If it is < or >, draw a solid line.
c. Pick any point {suggested (0,0), (0,1) or (1,0)} and see if it satisfies the inequality. If it does, shade that half of
the graph, if not, shade the other half of the graph.
Example: Graph: 2x + 3y < 6
Section 3.5 Introduction to Functions and Function Notation
Definitions:
Relation: Set of ordered pairs
Domain: Set containing initial (first) values of a relation; the initial values
Range: Set containing all (second) values that are paired to the domain values in the relation
Function: Relation in which each value in the domain is assigned to exactly one value in the range
There are three primary methods to define functions
Ordered Pairs { (x,y) | where x is the independent variable and y is the dependent variable.
Graphically:
y
f(x)=0.1x^3+1
x
y
f(x)=2x+4
x
y
f(x)=sin(x)+x
x
Equations: y = x2 -2 ; f(x) = 3x + 5
Ordered pairs: (independent variable, dependent variable)
The domain is the set of independent variables.
The range is the set of dependent variables.
Ex1. Determine the domain and range of each set and determine if it is a function.
16. { (8,-3), (1,5), (-2,6), (5,4) }
18. { (3,1), (4,-5), (4,-2), (3,-3) }
Graphs
The domain is the set containing the 1st coordinates (x axis)
The range is the set containing the 2nd coordinates (y axis)
Vertical Line Test
Look at any vertical line going thru the domain, if the line goes thru more than one point, it is not a function.
Ex2. Determine the domain and range of each graph and determine if it is a function.
24.
11 y
10
9
8
7
6
5
4
3
2
1
-11-10-9 -8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
f(x)=2-x
f(x)=x-2
x
1 2 3 4 5 6 7 8 9 10 11
28
11 y
10
9
8
7
6
5
4
3
2
1
-11-10-9 -8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
x
1 2 3 4 5 6 7 8 9 10 11
f(x)=2
f(x)=1.6667x-0.6667
f(x)=0.5x+3
Equations:
Given a function f(x), to find f(a) where a is a real value in the domain of f, replace x in the function with a and
then evaluate and simplify.
Ex3. 32. f(x) = -6x + 2; Find a) f(1), b) f(0), c) f(-1), d) f(t+3), x) f(x-2)
36. f(x) =
4x
2 ; a) f(-1), b) f(1.5), c) f(3), d) f(a), e) f(x+2)
Graphing Functions:
Linear functions are similar to linear equations in slope intercept form. e.g. f(x) = 3x – 2
Non-linear functions are not straight lines. e.g. f(x) = x2, f(x) = x3 - 1
Chapter 3 Review
<3.2>
7) What is the formula to determine the slope of a line between two distinct points?
16) Is (4/5, 2) a solution of 5x + y = 6
Graph:
20) y = 5x 24) 3x + 4y = 28
28) Determine the slope and y-intercept, then graph y – 4x + 3 = 0
Find the slope through the given points:
30) (3, -1), (3, 2)
31) (7, -4), (2, -9)
32) (-1, -1), (3, -1)
34) Write the equation in slope-intercept form given m = -2/5 going through (0, 4)
<3.3>
Write the equation in slope-intercept form going through
40) (-2, 2), (-4, -5)
42) (6, 0), (0, 6)
Determine if the graphs of the equations are parallel, perpendicular, or neither
43) y = 2/3 x ; y = 2/3 x - 4
44) 7x + y = -9; 7x – y = -6
46) x + y = -1; x – y = 5
48) Write the equation in slope-intercept form with a slope of 2 going through (-2, -2)
52) Find the equation of the line passing through (2, 5) that is parallel to x – 5y = 10
54) Find the equation of the line passing through (-3, -5) that is perpendicular to y – 3x = 4
<3.5>
**) Find the indicated value of the function f(x) = (x – 3)/(x + 5 ) for a) f(2) b) f(0)
c) f(-5) d) f(3)