If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra (all content)>Unit 13

Lesson 9: Discontinuities of rational functions

# Analyzing vertical asymptotes of rational functions

Sal analyzes the behavior of q(x)=(x²+3x+2)/(x+3) around its vertical asymptote at x=-3.

## Want to join the conversation?

• Wait a minute, q approaches +infinity when x-values are between -3 and -2. If x -> 3^+ means "values over -3," then why do we limit our values?
• It does not mean values over -3. It means the limit is approaching from the right side of 3. Most of the times, they are just same as when you put 3, but sometimes they are not the same.
• is simplifying the expressions eqivalent when i search for the asymptotes and their behaviour?
• When you simplify by cancelling, "(x-a)/(x-a)" = 1 for all x, so cancelling doesn't affect the graph - it won't affect any asymptotes that are left after the cancelling, or remove any asymptotes.
• I didn't quite get the answer of the first question. Shouldn't q(x) be approaching to positive infinity? I mean even Sal points that out on the numberline in blue color with the arrow pointing to the right side. What's happening... I'm confused.
• He starts to explain around , as q(x) approaches the vertical asymptote of -3, the function goes down and approaches negative infinity.

Try substituting any value less than -3 for x, and you'll find the function always comes out as a negative. If we look at x = -4, for example, the numerator simplifies to (-3)(-2) = 6. The denominator simplifies to -4+3 = -1. The function as a whole then simplifies to q(x) = -6 for x = -4. You can try this for any x value smaller than -3 and you'll find the function approaches negative infinity the closer x gets to it's vertical asymptote of -3.
• I would definitely start this video by considering the interval from -3 to -2 first (approaching from positive side), because if we start considering approaching from the negative, there is a temptation to use whole numbers such as (-5) and (-4) and by substituting these numbers (instead of -3.1 and -3.01) function DOES NOT appear to approach negative infinity. There is a sign change at x=-4.4.
• Could someone give me an example of a function where the asymptote approaches positive infinity from both directions or negative infinity from both directions?
• as x approaches positive infinity from left
f(x) approaches positive infinity from the right
(1 vote)
• How can we differentiate between horizontal and vertical asymptotes? Sometimes I hear Sal refer to horizontal asymptotes but like in this video, it focuses on vertical.
(1 vote)
• Horizontal asymptotes are when a function's y value starts to converge toward something as its x value goes toward positive or negative infinity. This is the end behavior of the function.

Vertical asymptotes are when a function's y value goes to positive or negative infinity as the x value goes toward something finite.

Let's say you have the function
a(x) = (2x+1)/(x-1).

As x → 1 from the negative direction, a(x) → -∞. As x → 1 from the positive direction, a(x) → +∞. This is your vertical asymptote, because as x approaches something finite, a(x) approaches something infinite.

As x gets bigger and bigger (you can think of this as x → ∞, I don't know if you have done end behavior at this point in the course), a(x) goes to 2. You can confirm this by plugging in really big values, for example:
(2(1000000000)+1)/((1000000000)-1) = 2.000000003
Anyway, the point is that as x approaches something infinite, a(x) approaches something finite, so this is your horizontal asymptote.

This is your bottom line:
Vertical asymptote: As x → finite, y → infinite
Horizontal asymptote: As y → finite, x → infinite

Hope this clarified a bit.
• So why do you have to restrict the -3^+ but not -3^-?
• You don't. You can approach -3 from either side. Just get close enough that y changes rapidly.
• Why do we care about the intervals -3<x<-2? What does Sal mean at when he says "I don't want any kind of weird sign changes"?
• Because, at x=-2, that would be a zero, and that has no sign. Also, if you stray too far from -3, you might get some wrong answers. Usually, I would check for numbers about 0.1 away from the asymptote maximum. The closer to the asymptote, the more accurate.
• In the example in the video, the line x = -3 is the vertical asymptote because it is the denominator and as x approaches -3 from either side, the denominator gets smaller at a faster rate than the rate at which the numerator is getting bigger.

If the factor (x+3) is also in the numerator, would this mean that x = -3 is no longer the vertical asymptote? Because now the numerator is changing at a pace that is just as fast as the denominator?