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Over [174]
2 years ago
13

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

point (0,3,5)
Mathematics
1 answer:
madam [21]2 years ago
3 0

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

Read more about exponential equation at

brainly.com/question/23729449

#SPJ1

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