Exponential function help.
$begingroup$
I have the following problem:
Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are 𝐴=(−1,15) and B=(1,6). The red horizontal line is given by y=3 and is a horizontal asymptote of the function.
Can someone show me how to solve this problem?
Thanks
exponential-function graphing-functions
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
$endgroup$
add a comment |
$begingroup$
I have the following problem:
Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are 𝐴=(−1,15) and B=(1,6). The red horizontal line is given by y=3 and is a horizontal asymptote of the function.
Can someone show me how to solve this problem?
Thanks
exponential-function graphing-functions
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
$endgroup$
add a comment |
$begingroup$
I have the following problem:
Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are 𝐴=(−1,15) and B=(1,6). The red horizontal line is given by y=3 and is a horizontal asymptote of the function.
Can someone show me how to solve this problem?
Thanks
exponential-function graphing-functions
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
$endgroup$
I have the following problem:
Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are 𝐴=(−1,15) and B=(1,6). The red horizontal line is given by y=3 and is a horizontal asymptote of the function.
Can someone show me how to solve this problem?
Thanks
exponential-function graphing-functions
exponential-function graphing-functions
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
asked Jan 21 at 0:07
NathanMNdz2NathanMNdz2
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume that $f(x)$ has the following form:
$$f(x) = a^{-bx +c} +3$$
The $+3$ is the necessary vertical offset for this function to have a horizontal asymptote at $y = 3$. The exponential function has a generally decreasing form which is required from the two points given.
We can then solve for $a$ and $b$ by using the solutions to $f(1)$ and $f(-1)$ as follows:
begin{align*}
a^{-b(1) + c} + 3 &= 6 \
-b + c &= log_a(3) \
c &= log_a(3) + b
end{align*}
begin{align*}
a^{-b(-1) + c} + 3 &= 15 \
b + c &= log_a(12) \
c &= log_a(12) - b
end{align*}
Using these two equations, we can find the following solutions:
begin{align*}
b &= log_a(2) \
c &= log_a(6)
end{align*}
So, we have the following:
$$f(x) = a^{-log_a(2)x +log_a(6)} +3$$
Which can be simplified in the following way:
begin{align*}
f(x) &= a^{-log_a(2)x +log_a(6)} +3 \
f(x) &= (a^{-log_a(2)})^xa^{log_a(6)} +3 \
f(x) &= (a^{log_a(frac{1}{2})})^xa^{log_a(6)} +3 \
f(x) &= left( frac{1}{2}right)^x(6) +3 \
f(x) &= 6left( frac{1}{2}right)^x +3 \
end{align*}
answered Jan 21 at 0:36
GustavGustav
1469
$endgroup$
add a comment |
$begingroup$
Substitute the points into the standard equation of the exponential function
$$mbox{$f(x) = amathrm{b}^x + k$}$$
You already know k = 3, so you get:
$$15 = amathrm{b}^{-1} + 3 ①$$
$$mbox{6 = ab + 3 --> b = $frac 3a$ ②}$$
sub ② into ①
$$mbox{12 = $frac {mathrm{a}^2}3$}$$
When a = 6, b = $frac 12$, When a = -6, b = -$frac 12$
Answer:
$$mbox{y = 6$mathrm{(frac 12)}^x$+ 3}$$
$$mbox{or}$$
$$mbox{y = -6$mathrm{(-frac 12)}^x$+ 3}$$
$$mbox{are both correct}$$
answered Jan 21 at 0:34
KevinKevin
326
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Assume that $f(x)$ has the following form:
$$f(x) = a^{-bx +c} +3$$
The $+3$ is the necessary vertical offset for this function to have a horizontal asymptote at $y = 3$. The exponential function has a generally decreasing form which is required from the two points given.
We can then solve for $a$ and $b$ by using the solutions to $f(1)$ and $f(-1)$ as follows:
begin{align*}
a^{-b(1) + c} + 3 &= 6 \
-b + c &= log_a(3) \
c &= log_a(3) + b
end{align*}
begin{align*}
a^{-b(-1) + c} + 3 &= 15 \
b + c &= log_a(12) \
c &= log_a(12) - b
end{align*}
Using these two equations, we can find the following solutions:
begin{align*}
b &= log_a(2) \
c &= log_a(6)
end{align*}
So, we have the following:
$$f(x) = a^{-log_a(2)x +log_a(6)} +3$$
Which can be simplified in the following way:
begin{align*}
f(x) &= a^{-log_a(2)x +log_a(6)} +3 \
f(x) &= (a^{-log_a(2)})^xa^{log_a(6)} +3 \
f(x) &= (a^{log_a(frac{1}{2})})^xa^{log_a(6)} +3 \
f(x) &= left( frac{1}{2}right)^x(6) +3 \
f(x) &= 6left( frac{1}{2}right)^x +3 \
end{align*}
answered Jan 21 at 0:36
GustavGustav
1469
$endgroup$
add a comment |
$begingroup$
Assume that $f(x)$ has the following form:
$$f(x) = a^{-bx +c} +3$$
The $+3$ is the necessary vertical offset for this function to have a horizontal asymptote at $y = 3$. The exponential function has a generally decreasing form which is required from the two points given.
We can then solve for $a$ and $b$ by using the solutions to $f(1)$ and $f(-1)$ as follows:
begin{align*}
a^{-b(1) + c} + 3 &= 6 \
-b + c &= log_a(3) \
c &= log_a(3) + b
end{align*}
begin{align*}
a^{-b(-1) + c} + 3 &= 15 \
b + c &= log_a(12) \
c &= log_a(12) - b
end{align*}
Using these two equations, we can find the following solutions:
begin{align*}
b &= log_a(2) \
c &= log_a(6)
end{align*}
So, we have the following:
$$f(x) = a^{-log_a(2)x +log_a(6)} +3$$
Which can be simplified in the following way:
begin{align*}
f(x) &= a^{-log_a(2)x +log_a(6)} +3 \
f(x) &= (a^{-log_a(2)})^xa^{log_a(6)} +3 \
f(x) &= (a^{log_a(frac{1}{2})})^xa^{log_a(6)} +3 \
f(x) &= left( frac{1}{2}right)^x(6) +3 \
f(x) &= 6left( frac{1}{2}right)^x +3 \
end{align*}
answered Jan 21 at 0:36
GustavGustav
1469
$endgroup$
add a comment |
$begingroup$
Assume that $f(x)$ has the following form:
$$f(x) = a^{-bx +c} +3$$
The $+3$ is the necessary vertical offset for this function to have a horizontal asymptote at $y = 3$. The exponential function has a generally decreasing form which is required from the two points given.
We can then solve for $a$ and $b$ by using the solutions to $f(1)$ and $f(-1)$ as follows:
begin{align*}
a^{-b(1) + c} + 3 &= 6 \
-b + c &= log_a(3) \
c &= log_a(3) + b
end{align*}
begin{align*}
a^{-b(-1) + c} + 3 &= 15 \
b + c &= log_a(12) \
c &= log_a(12) - b
end{align*}
Using these two equations, we can find the following solutions:
begin{align*}
b &= log_a(2) \
c &= log_a(6)
end{align*}
So, we have the following:
$$f(x) = a^{-log_a(2)x +log_a(6)} +3$$
Which can be simplified in the following way:
begin{align*}
f(x) &= a^{-log_a(2)x +log_a(6)} +3 \
f(x) &= (a^{-log_a(2)})^xa^{log_a(6)} +3 \
f(x) &= (a^{log_a(frac{1}{2})})^xa^{log_a(6)} +3 \
f(x) &= left( frac{1}{2}right)^x(6) +3 \
f(x) &= 6left( frac{1}{2}right)^x +3 \
end{align*}
answered Jan 21 at 0:36
GustavGustav
1469
$endgroup$
Assume that $f(x)$ has the following form:
$$f(x) = a^{-bx +c} +3$$
The $+3$ is the necessary vertical offset for this function to have a horizontal asymptote at $y = 3$. The exponential function has a generally decreasing form which is required from the two points given.
We can then solve for $a$ and $b$ by using the solutions to $f(1)$ and $f(-1)$ as follows:
begin{align*}
a^{-b(1) + c} + 3 &= 6 \
-b + c &= log_a(3) \
c &= log_a(3) + b
end{align*}
begin{align*}
a^{-b(-1) + c} + 3 &= 15 \
b + c &= log_a(12) \
c &= log_a(12) - b
end{align*}
Using these two equations, we can find the following solutions:
begin{align*}
b &= log_a(2) \
c &= log_a(6)
end{align*}
So, we have the following:
$$f(x) = a^{-log_a(2)x +log_a(6)} +3$$
Which can be simplified in the following way:
begin{align*}
f(x) &= a^{-log_a(2)x +log_a(6)} +3 \
f(x) &= (a^{-log_a(2)})^xa^{log_a(6)} +3 \
f(x) &= (a^{log_a(frac{1}{2})})^xa^{log_a(6)} +3 \
f(x) &= left( frac{1}{2}right)^x(6) +3 \
f(x) &= 6left( frac{1}{2}right)^x +3 \
end{align*}
answered Jan 21 at 0:36
GustavGustav
1469
edited Jan 21 at 1:01
answered Jan 21 at 0:36
GustavGustav
1469
answered Jan 21 at 0:36
GustavGustav
1469
answered Jan 21 at 0:36
GustavGustav
1469
1469
add a comment |
add a comment |
$begingroup$
Substitute the points into the standard equation of the exponential function
$$mbox{$f(x) = amathrm{b}^x + k$}$$
You already know k = 3, so you get:
$$15 = amathrm{b}^{-1} + 3 ①$$
$$mbox{6 = ab + 3 --> b = $frac 3a$ ②}$$
sub ② into ①
$$mbox{12 = $frac {mathrm{a}^2}3$}$$
When a = 6, b = $frac 12$, When a = -6, b = -$frac 12$
Answer:
$$mbox{y = 6$mathrm{(frac 12)}^x$+ 3}$$
$$mbox{or}$$
$$mbox{y = -6$mathrm{(-frac 12)}^x$+ 3}$$
$$mbox{are both correct}$$
answered Jan 21 at 0:34
KevinKevin
326
$endgroup$
add a comment |
$begingroup$
Substitute the points into the standard equation of the exponential function
$$mbox{$f(x) = amathrm{b}^x + k$}$$
You already know k = 3, so you get:
$$15 = amathrm{b}^{-1} + 3 ①$$
$$mbox{6 = ab + 3 --> b = $frac 3a$ ②}$$
sub ② into ①
$$mbox{12 = $frac {mathrm{a}^2}3$}$$
When a = 6, b = $frac 12$, When a = -6, b = -$frac 12$
Answer:
$$mbox{y = 6$mathrm{(frac 12)}^x$+ 3}$$
$$mbox{or}$$
$$mbox{y = -6$mathrm{(-frac 12)}^x$+ 3}$$
$$mbox{are both correct}$$
answered Jan 21 at 0:34
KevinKevin
326
$endgroup$
add a comment |
$begingroup$
Substitute the points into the standard equation of the exponential function
$$mbox{$f(x) = amathrm{b}^x + k$}$$
You already know k = 3, so you get:
$$15 = amathrm{b}^{-1} + 3 ①$$
$$mbox{6 = ab + 3 --> b = $frac 3a$ ②}$$
sub ② into ①
$$mbox{12 = $frac {mathrm{a}^2}3$}$$
When a = 6, b = $frac 12$, When a = -6, b = -$frac 12$
Answer:
$$mbox{y = 6$mathrm{(frac 12)}^x$+ 3}$$
$$mbox{or}$$
$$mbox{y = -6$mathrm{(-frac 12)}^x$+ 3}$$
$$mbox{are both correct}$$
answered Jan 21 at 0:34
KevinKevin
326
$endgroup$
Substitute the points into the standard equation of the exponential function
$$mbox{$f(x) = amathrm{b}^x + k$}$$
You already know k = 3, so you get:
$$15 = amathrm{b}^{-1} + 3 ①$$
$$mbox{6 = ab + 3 --> b = $frac 3a$ ②}$$
sub ② into ①
$$mbox{12 = $frac {mathrm{a}^2}3$}$$
When a = 6, b = $frac 12$, When a = -6, b = -$frac 12$
Answer:
$$mbox{y = 6$mathrm{(frac 12)}^x$+ 3}$$
$$mbox{or}$$
$$mbox{y = -6$mathrm{(-frac 12)}^x$+ 3}$$
$$mbox{are both correct}$$
answered Jan 21 at 0:34
KevinKevin
326
edited Jan 21 at 1:14
answered Jan 21 at 0:34
KevinKevin
326
answered Jan 21 at 0:34
KevinKevin
326
answered Jan 21 at 0:34
KevinKevin
326
326
add a comment |
add a comment |
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