Question

How do I find the vertical asymptote of
f(x) = 3x+2 / x-5 ?

Answer 1

Do you mean f(x) = (3x+2)/(x-5)? It makes a HUGE difference.

You have written \(f(x) = 3x + \dfrac{2}{x} - 5\) and the vertical asymptote is x = 0.

You may have meant \(f(x) = \dfrac{3x+2}{x-5}\) and the vertical asymptote is x = 5.

Please remember your Order of Operations and use parentheses to clarify intent.

Answer 2

The second one.

Answer 3

Thank you. Is 5 also the domain?

Answer 4

No. Vertical Asymptotes are not included in the Domain. Everything except x = 5.

Did you learn your lesson about the Order of Operations? It's important for clear communication.

Answer 5

yes I did learn about the order of operations and how to communicate it. Thank you. So how do I find the domain and horizontal asymptote?

Answer 6

Domain is trivial. All real numbers except x = 5. Done.

An horizontal asymptote requires a little judgment. You must examine the DEGREE of the numerator and denominator. Tell me what those are.

Answer 7

2nd degree?

Answer 8

I do not understand the degree of the numerator or denominator

Answer 9

No good. Both have only 'x'. There is no '\(x^{2}\)'

Look for the highest exponent on any one term in the expression.

(3x+2) -- Highest exponent is 1 (\(x^{1} = x\)) - This is 1st degree or linear.

(x-5) -- Highest exponent is 1 - This is 1st degree or linear.

Do you believe?

Answer 10

yes. thank you, so what do I do next?

Answer 11

OTHER asymptotes come from these degrees.

If the degree of the numerator is LESS than the degree of the denominator, there is an horizontal asymptote and it is y = 0.

If the degree of the numerator is EQUAL to the degree of the denominator, there is an horizontal asymptote and it is y = the ratio of the leading coefficients. In your case, it is y = (3/1) or just y = 3. Do you see that this is so?

If the degree of the numerator is GREATER than the degree of the denominator, there is NOT an horizontal asymptote. Depending on the difference, it could be a line, a parabola, or lots of other things.

Answer 12

I do understand it better now. For the problem...would the critical numbers = 3, 5...or just 5?

Answer 13

Hmm... Depends on the definition of "Critical Numbers". That would normally be be x-values that are significant for some reason.

x = 5 is one, because there is an asymptote there.
x = -2/3 is one, because that is an x-intercept. Do you see this? (Substitute f(x) = 0)

y = -2/5 might be one, because that is the y-intercept. Do you see this? (Substitute x = 0)
y = 3 might be one, because there is an horizontal asymptote there.

You'll have to check your course materials or class notes to see exactly which ones are intended.

gtg - Good luck!