Finding an Exponential Function Algebraically - Purdue University
[Pages:6]16-week Lesson 30 (8-week Lesson 24)
Finding an Exponential Function Algebraically
In this set of notes we'll be covering finding exponential functions
algebraically by using ordered pairs that the graph of the function passes
through. We'll use a similar process to how we found quadratic functions
from ordered pairs in Lesson 24. In Lesson 24 we were always given the vertex of the quadratic function, and we'd simply plug that in for and . Then we'd use the other ordered pair we were given to find the value of . With exponential functions, () = , we will always be given the -intercept of the function and we'll simply plug that in for . Then we'll use the other ordered pair we're given to find the value of the base .
Example 1: Find an exponential function of the form () = , given that the graph of the function passes through the points (0, 3) and (1,2). Simplify your answer completely and enter exact answers only
Always start by plugging in the -intercept in order to find the value of .
() = (0) = 0
3 = 1
The ordered pair (0, 3) says that when 0 is the input of the function , 3 is the output. That means that (0) = 3. And since
those two values are equivalent,
they are also interchangeable.
Now that we know that value of is 3, we can replace with 3 in our exponential function.
() = 3
Now use the other ordered pair (1, 2) to find the value of .
(1) = 3 1 2 = 3 2 3 =
The ordered pair (1,2) says that when 1 is the input of the function , 2 is the output. That means that (1) = 2. And since
those two values are equivalent,
they are also interchangeable.
Now
that
we
know
that
value
of
is
23,
we
can
replace
with
2 3
in
our
exponential function.
()
=
()
1
16-week Lesson 30 (8-week Lesson 24)
Finding an Exponential Function Algebraically
Always start by plugging in the -intercept to find the value of , then use the other ordered pair to find the value of .
Example 2: Find an exponential function of the form () = , given that the graph of the function passes through the points (0, 2) and (-2, 50). Simplify your answer completely and enter exact answers
only (no approximations).
Remember from the notes on Exponential Functions in Lesson 29 that for any exponential function () = , the base must be a positive number other then one ( > 0, 1). So when you encounter an value such as ? 15, be sure to disregard the negative value.
2
16-week Lesson 30 (8-week Lesson 24)
Finding an Exponential Function Algebraically
Example 3: Find an exponential function of the form () = , given
that the -intercept of the function is 1, and the graph of the function passes through the point (-1, 45). Simplify your answer completely and enter exact answers only (no approximations).
() =
() =
(0) = 0
(-1) = -1
1 = 1
4=1
5
= 1
4 = 5
() = 1
= 5
4
() =
() = ()
Example 4: Find an exponential function of the form () = , given that the -intercept of the function is - 13, and the graph of the function passes through the point (3, - 3433). Simplify your answer completely
and enter exact answers only (no approximations).
() =
(0) = 0
- 1 = 1
3
= - 1
3
()
=
-
1 3
() =
() = - 1
3
(3) = - 1 3
3
- 343 = - 1 3
3
3
343 = 3
7 =
() = -
3
16-week Lesson 30 (8-week Lesson 24)
Finding an Exponential Function Algebraically
Example 5: Find an exponential function of the form () = , given
that
the
graph
of
the
function
passes
through
the
points
(0,
3)
2
and
(-1, 3). Simplify your answer completely and enter exact answers only
(no approximations).
Example 6: Find an exponential function of the form () = , given that the graph of the function passes through the points (0, -1) and (2, -2). Simplify your answer completely and enter exact answers only (no approximations).
4
16-week Lesson 30 (8-week Lesson 24)
Finding an Exponential Function Algebraically
Example 7: Find an exponential function of the form () = , given
that the -intercept of the function is -2, and the graph of the function passes through the point (-2, -22). Simplify your answer completely
and enter exact answers only (no approximations).
Always start by plugging in the -intercept in order to find the value of .
In this case we need to express the -intecept of -2 as the ordered pair (0, -2).
() =
(0) = 0
-2 = 1
Now that we know the value of , we can replace with -2.
() = -2
Now use the other ordered pair we were given to find the value of .
(-2) = -2 -2
-22 = -2 -2
2
=
1 2
2 2 = 1
2
=
1 2
= ?12
1 = ? Since the base of an exponential function must be a positive number other than 1, we drop the negative root and keep only the positive root
() = - ()
5
16-week Lesson 30 (8-week Lesson 24)
Finding an Exponential Function Algebraically
Answers to Examples:
1.
()
=
3
(2)
3
;
2.
()
=
2
(1)
5
;
3.
()
=
(5)
4
;
4.
() = - 1 (7)
3
;
5.
()
=
3 2
()
2
;
6.
() = -;
7. () = -2 (1);
6
................
................
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