CHAPTER 1:
Graphs, Functions,
and Models
1.1 Introduction to Graphing
1.2 Functions and Graphs
1.3 Linear Functions, Slope, and Applications 1.4 Equations of Lines and Modeling
1.1
Introduction to Graphing
Plot points.
Determine whether an ordered pair is a solution of an
equation.
Find the x-and y-intercepts of an equation of the form
Ax + By = C.
Graph equations.
Find the distance between two points in the plane and find the
midpoint of a segment.
Find an equation of a circle with a given center and radius, and
given an equation of a circle in standard form, find the center and the radius.
To graph or plot a point, the first coordinate tells us to move left or right from the origin. The second
coordinate tells us to move up or down.
Plot (3, 5).
Move 3 units left.
Next, we move 5 units up. Plot the point.
Solutions of Equations
Equations in two variables have solutions (x, y) that are ordered pairs.
Example: 2x + 3y = 18
Examples
a.
Determine whether the ordered pair (5, 7) is a solution of 2x + 3y = 18.
2(5) + 3(7) ? 18
10 + 21 ? 18 11 = 18 FALSE
(5, 7) is not a solution.
b.
Determine whether the ordered pair (3, 4) is a solution of 2x + 3y = 18.
2(3) + 3(4) ? 18 6 + 12 ? 18 18 = 18 TRUE
Graphs of Equations
To graph an equation is to make a drawing that
x
-Intercept
The point at which the graph crosses the x-axis.
An x-intercept is a point (a, 0). To find a, let y = 0 and solve for x.
Example: Find the x-intercept of 2x + 3y = 18. 2x + 3(0) = 18
2x = 18 x = 9
y
-Intercept
The point at which the graph crosses the y-axis.
A y-intercept is a point (0, b). To find b, let x = 0 and
solve for y.
Example: Find the y-intercept of 2x + 3y = 18. 2(0) + 3y = 18
3y = 18 y = 6
Example
We already found the x-intercept: (9, 0)
We already found the y-intercept: (0, 6)
We find a third solution as a check. If x is replaced with 5, then
Graph 2x + 3y = 18.
2 5 3y 18
10 3y 18
3y 8
y 8
3
Thus, is a solution.5, 8
Example
(continued) Graph:2x + 3y = 18.
x-intercept:
(9, 0)
y-intercept:
(0, 6)
Third point:
5, 8 3
Example
Graph y = x2 – 9x – 12 .
(12, 24) 24 12 –2 32 32 26 12 –2 24 y (10, –2) 10
(5, 32) 5
(4, 32) 4
(2, 26) 2
(0, 12) 0
(1, –2)
1
(3, 24)
3
(x, y)
x
The Pythagorean Theorem
a
b
c
a
2+
b
2=
c
2Which side is
the hypotenuse?
a
2=
c
2–
b
2The right angle points to the hypotenuse.
6
8
c
Calculate side c.
c
2=
8
2+ 6
2c
2= 64
+ 36
c
2= 100
100
c
c
= 10
When calculating the
hypotenuse, we add the area of the squares of the other two sides.
Tanya is making a party hat using a
cone made out of paper. Determine
the height of the cone.
b
2=
c
2–
a
2h
2= 144
h
= 12 cm
h
2= 13
2– 5
2h
2= 169– 25
144
h
h
5 cm
The Distance Formula
The distance d between any two points
(x1, y1) and (x2, y2) is given by
d (x2 x1)2 (y
Example
Find the distance between (4,8) and (1,12)
2 2
2 1 2 1
distance
(
x
x
)
(
y
y
)
(4, 8)
(1, 12)
2 2
distance
(1 4)
(12 8)
2 2
distance
( 3)
(4)
Example
d (3
2 )2 (6 2)2d 52 (8)2 25 64
d 89 9.4
Midpoint Formula
If the endpoints of a segment are (x1, y1) and (x2, y2), then the coordinates of the midpoint are
x1 x2
2 ,
y1 y2
2
Example
Find the midpoint of a segment whose endpoints are (4, 2) and (2, 5).
4 2
2 ,
2 5
2
2
2 , 32
1, 3
Circles
A circle is the set of all points in a plane that are a
fixed distance r from a center (h, k).
The equation of a circle with center (h, k) and radius
r, in standard form, is
Example
Find an equation of a circle having radius 5 and center (3, 7).
Using the standard form, we have (x h)2 + (y k)2 = r2
[x 3]2 + [y (7)]2 = 52
Ex. 2: Writing a Standard Equation of a
Circle
The point (1, 2) is on a circle whose center is (5, -1). Write a standard equation of the circle.
2 1 2
2 1
2 ) ( )
(x x y y
r =
r = (5 1)2 (1 2)2 r = (4)2 (3)2
r = 16 9
r = 25
Use the Distance Formula Substitute values.
Simplify.
Simplify.
Addition Property
Ex. 2 (Cont’d): Writing a Standard
Equation of a Circle
The point (1, 2) is on a circle whose center is (5, -1). Write a standard equation of the circle.
(x – h)2 + (y – k)2 = r2 Standard equation of a circle.
[(x – 5)]2 + [y –(-1)]2 = 52 Substitute values.
