The exponential equation is [tex]y = \frac23(3)^{-x+0.74} + 2[/tex]
How to determine the equation?
An exponential function is represented as:
[tex]y = b^x[/tex]
The base is 3.
So, we have:
[tex]y = 3^x[/tex]
It is stretched vertically by 2/3.
So, we have:
[tex]y = \frac23(3)^x[/tex]
When reflected over the y-axis, we have:
[tex]y = \frac23(3)^{-x}[/tex]
An asymptote of y = 2, makes the function becomes
[tex]y = \frac23(3)^{-x} + 2[/tex]
Lastly, it passes through the point (0, 3.5).
So, we have:
[tex]y = \frac23(3)^{-x+h} + 2[/tex]
This gives
[tex]3.5 = \frac23(3)^{-0+h} + 2[/tex]
[tex]3.5 = \frac23(3)^{h} + 2[/tex]
Subtract 2 from both sides
[tex]1.5 = \frac23(3)^{h}[/tex]
Multiply by 3/2
[tex]2.25 = (3)^{h}[/tex]
Take the logarithm of both sides
log(2.25) = h * log(3)
Solve for h
h = 0.74
Substitute h = 0.74 in [tex]y = \frac23(3)^{-x+h} + 2[/tex]
[tex]y = \frac23(3)^{-x+0.74} + 2[/tex]
Hence, the exponential equation is [tex]y = \frac23(3)^{-x+0.74} + 2[/tex]
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