Question

What are the asymptote(s) and hole(s), if any, of `f(x)= (2-e^(x))/(3x-2xe^(x/2))`?

Answer 1

Vertical Asymptotes: x=0, `ln(9/4)`
Horiziontal Asymptotes: y = 0
Oblique Asymptotes: None
Holes: None

Explanation:

The `e^x` parts may be confusing but don't worry, just apply the same rules.
I'll start with the easy part: The Vertical Asymptotes
To solve for those you set the denominator equal to zero as a number over zero is undefined. So:
`3x-2xe^(x/2)=0`
Then we factor out an x
`x(3-2e^(x/2))=0`
So one of the vertical asymptotes is x = 0. So if we solve the next equation.
`(3-2e^(x/2))=0` Then use algebra, isolate the exponent: `-2e^(x/2)=-3`
Then divide by -2: `e^(x/2) = 3/2`
Finally, we take the natural log of both sides as a means of canceling out the exponent: `ln(e^(x/2))=ln(3/2)`
So on the left, we are left with `x/2 = ln(3/2)`
So this final zero is `x = 2 ln(3/2)` and because of exponent log property that states `ln(x^n) = n * ln(x)`, it is equivalent to `x = ln(9/4)`
So now that we've established that, the rest is easy. Because the numerator doesn't divide into the denominator, there can't be an oblique asymptote. Also, the denominator has a larger degree than the numerator. And when you try to factor the denominator, as shown above, none of the factors match the numerator
Finally, to close off, we have a horizontal asymptote of y=0 because the `e^x` function never equals zero.
Key Points:
1. `e^x ne 0`