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OpenStudy (anonymous):
What exponential function is the best fit for the data in the table?
x f(x)
1 -4
3 -1
4 3
f(x) = 4(3)x - 1 + 4
f(x) = 4(3)x - 1 - 4
f(x) = one fourth(3)x - 1 + 4
f(x) = one fourth(3)x - 1 - 4
12 years ago
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OpenStudy (anonymous):
i plugged in the x's but it just doesn't make sense
12 years ago
OpenStudy (anonymous):
f(x) = 4(3)^(x - 1) + 4
f(x) = 4(3)^(x - 1) - 4
f(x) = 1/4(3)^(x - 1) + 4
f(x) = 1/4(3)^(x - 1) - 4
12 years ago
OpenStudy (anonymous):
@wio @precal
12 years ago
OpenStudy (anonymous):
What subject is this?
12 years ago
OpenStudy (anonymous):
algebra 11
12 years ago
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OpenStudy (anonymous):
2
12 years ago
OpenStudy (anonymous):
First of all, the function is increasing, so the base of the exponent should be \(>1\) ok?
12 years ago
OpenStudy (anonymous):
sure
12 years ago
OpenStudy (anonymous):
All exponential functions can be of the form :\[
f(x) = b^x+c
\]
12 years ago
OpenStudy (anonymous):
In our case, we say \(b>1\).
12 years ago
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OpenStudy (anonymous):
but we already have our b
12 years ago
OpenStudy (anonymous):
dont we?
12 years ago
OpenStudy (anonymous):
whoops \[
f(x) = ab^x+c
\]
12 years ago
OpenStudy (anonymous):
What is \(b\)?
12 years ago
OpenStudy (anonymous):
i believe from the answer choices it would be 3
12 years ago
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OpenStudy (anonymous):
When \(x = 0\), then we have \[
ab^0+c = a+c
\]
12 years ago
OpenStudy (anonymous):
true
12 years ago
OpenStudy (anonymous):
It looks like they have \[
ab^{x-1}+c
\]So in this case, \(x=1\) would be where our \(x=0\) would have been.
12 years ago
OpenStudy (anonymous):
pretty much but it still does not match the chart
12 years ago
OpenStudy (anonymous):
What is closest?
12 years ago
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OpenStudy (anonymous):
ooooh am i suppose look for the closest?
12 years ago
OpenStudy (anonymous):
well the closest is D
12 years ago
OpenStudy (anonymous):
D gets -3.75
12 years ago
OpenStudy (anonymous):
hmmmm........
12 years ago
OpenStudy (anonymous):
nothing else actually works
A = 8
B = 0
C = 4.25
D = -3.75
12 years ago
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OpenStudy (anonymous):
@wio are you there
12 years ago
OpenStudy (anonymous):
well thanks for your help wio you helped me get a 100%
12 years ago
OpenStudy (anonymous):
legit
12 years ago
OpenStudy (anonymous):
gracias
12 years ago