4.1
Linear Functions: Graphs & Models
Homework: EOO 1-17, 20, EOO 21-33, 41, 52
| Nancy J. McCormick's Home Page | DSPM 0800 Class Notes |
A. SLOPE-INTERCEPT EQUATION:
y = mx +
b
or
f (x) = mx + b (function notation)
Using the grapher, graph the following.
y = 2x
y = 2x –
7
y = 2x + 5
What characteristic do you notice about the graphs? The
lines are PARALLEL. For these equations, the lines are all increasing
(the slope is positive) at the same rate. The lines all have the same SLOPE.
What number was the same in each of the three equations? The coefficient of x
was 2 in each equation.
We conclude that 2 is the SLOPE of each line.
* In general, for a linear function:
f (x) = mx + b
m, the coefficient of x , is the SLOPE
of the line.
Look again at the graphs of
y =
2x
y = 2x – 7
y = 2x + 5
Where does y = 2x cross the y-axis? At (0, 0)
Where does y = 2x -7 cross the y-axis? At (0, -7)
Where does y = 2x + 5 cross the y-axis? At (0,
5)
Can you find the numbers 0, -7,
and 5 in the equations? Notice that 0
is not written in the first equation because 2x +
0 is equal to 2x.
We conclude that the constant b in
each equation is the y-coordinate of the Y-INTERCEPT. (Recall that for
any y-intercept, the x-coordinate is 0).
* In general, for a linear function:
f (x) = mx + b
m, the coefficient of x , is the SLOPE of the line.
b, the constant , is the y-intercept of the line.
Example:
Find a linear function for a line with slope 2
and y-intercept (0, -1).
f(x) = mx + b
f(x) = 2x – 1
B. Given
the slope and the y-intercept, you can graph the line by hand.
Example:
Find a linear equation for a line with slope 2/5 and y-intercept (0, 4)
and graph the line using the y-intercept and the slope to get a second point
on the line.
y =
mx + b
Begin by plotting the y-intercept (0, 4). The slope is 2/5 which is a ratio of vertical change to horizontal change. Move up 2 units and right 5 units to arrive at a second point on the line. (0 + 5, 4 + 2) or (5, 6).
Notice that you can rename the slope as –2/-5 and from the y-intercept, move down 2 units and left 5 units (0 – 5, 4 – 2) or (-5, 2) to find a third point on the line.
