Find the equation of the logarithmic function that passes through (-2,2) and (2,0) that has a vertical asymptote at x=3?

Answer 1

The "parent" logarithm function is y = log(x). This has a vertical asymptote at x = 0.

If the asymptote has been moved to the right 3 spaces, that means that we have either

y = log(x-3) or y = log(3-x)

Because points exist to the left of x=3, we must assume that the logarithm function lies to the left of x=3, which makes it the y = log(3-x).

This is consistent with our point (2,0) because when x=2, y =log(3-2)=log(1)=0

But we have not accounted for our function passing through (-2,2), so we will modify our function slightly, and let y = A·log(3-x) Now plug in x=-2 and y=2 and find A.

2 = A⋅log(3-(-2))

2=A⋅log(5)

So A = 2/log(5)

Now our function is:

f(x) = [ 2/log(5) ] ⋅ log(3-x)

The video has a graph of it.