Graphs and
Graphing Utilities
Definitions
• The horizontal number line is the x-axis.
• The vertical number line is the y-axis.
• The point of intersection of these axes is
their zero points, called the origin.
5
4
3
Origin (0, 0)
2
1
-5 -4 -3 -2
-1
-1
-2
-3
-4
-5
1
2
3 4
5
Definitions
• The axes divide the plane into four quarters,
called quadrants.
• Each point in the rectangular coordinate
system corresponds to an ordered pair of
real numbers, (x, y).
5
2nd quadrant
4
3
1st quadrant
2
1
-5 -4 -3 -2
-1
-1
-2
3rd quadrant -3
1
2
3 4
5
4th quadrant
-4
-5
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Example
Plot the point (3,2).
Start at the origin and move 3 units to the
right.
From that point, move 2 units up.
Now plot your point.
Text Example
Sketch the graph of y = x2 – 4.
Let x = 3, then y = x2 – 4 = 9 – 4 = 5.
The ordered pair (3, 5) is a solution to the
equation y = x2 – 4.
We also say that (3, 5) satisfies the equation.
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Text Example Cont.
• First, find several ordered pairs that are
solutions to the equation.
x
y = x2 – 4
-3
y = (-3)2 – 4 = 9 – 4 = 5
-2
y = (-2)2 – 4 = 4 – 4 = 0
Ordered Pair (x, y)
(-3, 5)
(-2, 0)
y=
(-1)2
0
y=
(0)2
1
y = 12 – 4 = 1 – 4 = -3
(1, -3)
2
y = 22 – 4 = 4 – 4 = 0
(2, 0)
3
y = 32 – 4 = 9 – 4 = 5
(3, 5)
-1
(-1, -3)
– 4 = 1 – 4 = -3
– 4 = 0 – 4 = -4
(0, -4)
Text Example Cont.
• Now, we plot these ordered pairs as points
in the rectangular coordinate system.
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
1
2 3 4
5
-3
-4
-5
Example
Graph 4y + 5x = 20.
Substitute zero for x:
4y = 20 or y = 5.
Hence, the y-intercept is (0,5).
Substitute zero for the y:
5x = 20 or x = 4.
Hence, the x-intercept is (4,0).
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Graphs and
Graphing Utilities
Distance and Midpoint
Formulas
4
The Distance Formula
• The distance, d, between the points (x1, y1)
and (x2,y2) in the rectangular coordinate
system is
d = (x 2 − x1 )2 + (y2 − y1 ) 2
Example
Find the distance between (-1, 2) and (4, -3).
Solution Letting (x1, y1) = (-1, -3) and (x2, y2) = (2, 3), we obtain
d = (x 2 − x1 ) + (y2 − y1 )
2
2
d = (4 − (−1))2 + ((−3) − 2)2
d = (5)2 + (−5)2
d = 25 + 25
d = 50
d=5 2
The Midpoint Formula
• Consider a line segment whose endpoints
are (x1, y1) and (x2, y2). The coordinates of
the segment's midpoint are
 x1 + x2 y1 + y2 
,


 2
2 
• To find the midpoint, take the average of
the two x-coordinates and of the two ycoordinates.
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Text Example
Find the midpoint of the line segment with
endpoints (1, -6) and (-8, -4).
Solution
To find the coordinates of the midpoint, we
average the coordinates of the endpoints.
 1 + (−8) −6 + (−4)   −7 −10   −7

,

 = ,
 =  ,−5
 2





2
2 2
2
(-7/2, -5) is midway between the points (1, -6) and (-8, -4).
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