Question

f(x)=(x^2-9)/(x-3) and g(x)=(x-9)/(x-3)
both have discontinuities at x = 3. Describe what happens in each function at x = 3. Be specific in your description for each function.

Answer 1

So we have equations

\[f(x) = (x ^{2}-9)/ x-3 \] this could also be written as \[f(x) = (x-3)(x+3)/(x-3)\] we can simplify the function to \[f(x) = x+3\]

We can say that this function has a vertical asymptote at x= 3.
We can also say hat this function has a oblique asymptote of function y = x+3. You can find the asymptote by doing long division. So there is a hole at x = 3. The f(3)
co-ordinates are (3, 6).

The function \[g(x) = (x-9)/(x-3)\] has a vertical asymptote at x=3.
This function has no oblique asymptote. There is a horizontal asymptote at y = 0.
g(3) = undefined. Approaching the function from the right g(3.000001) we approach \[-\infty \] Approach the function from the left g(2.99999) we approach
\[+\infty \]



I hope this helps.... check yourself to confirm...

Answer 2

thank you so much!

Answer 3

yw