A function "increasing" when the 𝑦-value increases as
the 𝑥-value increases, like this:
A function “decreasing" when the 𝑦-value decreases
as the 𝑥-value increases, like this:
𝑓 𝑥 = 𝑥 2 − 2𝑥 − 3. To find its 𝑥-intercept(s). Set
𝑓 𝑥 = 0 and solve for 𝑥. Namely, 𝑥 2 − 2𝑥 − 3 =
0
𝑥 + 1 𝑥 − 3 = 0 𝑥 = −1 or 𝑥 = 3. Thus,
−1,0 and (3,0) are the 𝑥-intercepts
• The 𝑥-intercept is the point
at which the graph of an
function crosses the 𝑥-axis.
For a point to cross the xaxis it must have a 𝑦 value
of zero. As a result, an 𝑥intercept can always be
represented by 𝑥, 0 .
The 𝑥 value indicates where
the graph crosses the 𝑥-axis.
To find the 𝑥-intercept(s)
set 𝑦 or function equal to
zero and solve for 𝑥.
𝑓(𝑥) = 2𝑥 − 1. Let 𝑥 = 0, then 𝑓 0 = 2 ×
0 − 1 = −1, so (0, −1) is the 𝑦-intercept
• The 𝑦-intercept is the point
at which the graph of an
function crosses the 𝑦-axis.
For a point to cross the yaxis it must have a 𝑥 value
of zero. As a result, an 𝑦intercept can always be
represented by 0, 𝑦 .
The 𝑦 value indicates where
the graph crosses the 𝑦-axis.
To find the 𝑦-intercept
substitute 0 for 𝑥 and solve
for 𝑦.
Copyright © 2008 Pearson
Education, Inc. Publishing as
Pearson Addison-Wesley
The following functions give the populations of four towns with time t
in years
• A. 𝑃 = 212 1.16
• C. 𝑃 = 835 0.89
𝑡
𝑡
B. 𝑃 = 1045 1.08 𝑡
D. 𝑃 = 560 1.01 𝑡
• a) Which town has fast rate of growth
• A has the largest base (fast rate of growth).
The population of the town A grows by a
factor of 1.16 each year.
Find an exponential function fitting the data:
4.48
= 0.7 or 4.48 = 6.4 × 0.7 (second value= first value× 0.7)
6.4
3.136
= 0.7 or 3.136 = 4.48 × 0.7 (second value= first value× 0.7)
4.48
2.1952
= 0.7 or 2.1952 = 3.136 × 0.7 (second value= first value× 0.7)
3.136
𝑓 𝑥 = 𝐴𝑏 𝑥 , 𝐴 = 6.4 since 𝐴 = 𝑓(0) . 𝑏 = 0.7 since the value of 𝑓 𝑥 is multiplied
by 0.7 for every one-unit increase of 𝑥. So 𝑓 𝑥 = 6.4 0.7 𝑥
𝒙
𝟎
𝟏
𝟐
𝟑
𝑓(𝑥)
6.4
4.48
3.136
2.1952
Assume that 𝐴 is the initial value
• If an exponential function 𝑓 𝑥 increases by
𝑚% per unit ( 𝑓(𝑥) is multiplied by a factor
1 + 𝑚% for every one unit increase of 𝑥)
• Then 𝑓 𝑥 = 𝐴 1 + 𝑚% 𝑥
The amount of a hormone in the body can change rapidly.
Suppose the initial amount is 20 mg. Find a formula for 𝐻, the
amount in mg, at a time t minutes later if 𝐻 is
(b) Increasing by 3% per minute
𝐻 = 20 1 + 3% 𝑡 = 20 1 + 0.03
20 1.03 𝑡 mg
𝑡
=
• If an exponential function 𝑓 𝑥 dencreases by
𝑚% per unit ( 𝑓(𝑥) is multiplied by a factor
1 − 𝑚% for every one unit increase of 𝑥)
• Then 𝑓 𝑥 = 𝐴 1 − 𝑚% 𝑥
The amount of a hormone in the body can change rapidly.
Suppose the initial amount is 20 mg. Find a formula for 𝐻, the
amount in mg, at a time t minutes later if 𝐻 is
(d) decreasing by 3% per minute
𝐻 = 20 1 − 3% 𝑡 = 20 1 − 0.03
20 0.97 𝑡 mg
𝑡
=