write equation that has base of 3 stretched vertically by factor of 2/3 reflected in y axis, asymptote of y=2 and passes through

Question

2 years ago

13

point (0,3,5)

1 answer:

madam [21] · Answer 1 · 2023-07-06T21:03:25+03:00

The exponential equation is $y = \frac23(3)^{-x+0.74} + 2$

<h3>How to determine the equation?</h3>

An exponential function is represented as:

$y = b^x$

The base is 3.

So, we have:

$y = 3^x$

It is stretched vertically by 2/3.

So, we have:

$y = \frac23(3)^x$

When reflected over the y-axis, we have:

$y = \frac23(3)^{-x}$

An asymptote of y = 2, makes the function becomes

$y = \frac23(3)^{-x} + 2$

Lastly, it passes through the point (0, 3.5).

So, we have:

$y = \frac23(3)^{-x+h} + 2$

This gives

$3.5 = \frac23(3)^{-0+h} + 2$

$3.5 = \frac23(3)^{h} + 2$

Subtract 2 from both sides

$1.5 = \frac23(3)^{h}$

Multiply by 3/2

$2.25 = (3)^{h}$

Take the logarithm of both sides

log(2.25) = h * log(3)

Solve for h

h = 0.74

Substitute h = 0.74 in $y = \frac23(3)^{-x+h} + 2$

$y = \frac23(3)^{-x+0.74} + 2$

Hence, the exponential equation is $y = \frac23(3)^{-x+0.74} + 2$

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