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### Course: Precalculus>Unit 4

Lesson 4: Graphs of rational functions

# Graphs of rational functions: vertical asymptotes

Sal picks the graph that matches f(x)=g(x)/(x²-x-6) (where g(x) is a polynomial) based on its discontinuities.

## Want to join the conversation?

• How can we be sure of what the question requires?like isn't there a way to figure out whether a function leads to the formation of a vertical asymptote or when it would lead to a discontinuity in the graph? For eg. in this ques. you said it could either be a vertical asymptote or a discontinuity.Isn't there a definite way out...so that we can look out for that particular thing itself.
• This example is a question about interpreting the parts of expressions. At first, rational functions seem wildly complicated. It isn't like the equation of a line, (linear function), f(x) = mx +b, where you just have slope and intercept to worry about. With a little practice, though, you can figure out a lot about a graph by looking at the parts of these rational functions.
In this case, the `only detail` we have is that there is a quadratic in the denominator. We have no information about the numerator. The denominator of a rational function can't tell you about the horizontal asymptote, but it CAN tell you about possible vertical asymptotes. What Sal is saying is that the factored denominator (x-3)(x+2) tells us that either one of these would force the denominator to become zero -- if x = +3 or x = -2. If the denominator becomes zero then the function is undefined at THAT point -- that value of x.
So, what happens when the denominator of a rational function becomes zero? It either just has a gap at that point (indicated by an empty dot, or little circle), or it has a vertical asymptote at that point which means that the function swoops toward positive or negative infinity, never crossing that point. `IF` we could see the numerator, we could find out whether each of the binomial factors cancels out with a corresponding binomial factor in the numerator. If so, it would be an empty circle on the graph (shows up as ERROR on table list of calculator), which means it is a `removable discontinuity`. If the binomial factor remains in the denominator because it cannot be cancelled, it will show up as a `vertical asymptote` on the graph at the value of x that would be undefined.

So, to answer your final question, in this specific example, we cannot tell `which` would happen without seeing the numerator. Instead we should look for `either` two asymptotes at the correct spots `or` two empty dots at the correct spots `or` one of each.
• How did he determine that 3 was the removable discontinuity?
• Sal checked what was happening at x = -2 and at x = 3.
At x = -2, the dotted line indicates an asymptote,, a line that the graph does not cross. But a removable discontinuity is a single point that cannot be included.
The open circle at x = 3 indicates that that specific point is not included. Therefore, that is the removable discontinuity.
• Around , Sal talks about x=3 possibly being a removable discontinuity. What about x=-2? We do not know the numerator ( g(x) ). So why doesn't Sal mention x=-2 possibly being a removable discontinuity? Could a possible graph show x=-2 as a removable discontinuity and x=3 as a vertical asymptote?
• Yes, x = -2 can be a removable discontinuity, but in this problem, we need to ensure both x = -2 and x = 3 is undefined. This is because the factored form of the denominator is (x-3)(x+2). Regardless of being a vertical asymptote or a removable discontinuity, both values of x (making the denominator 0) must be undefined. Wheter x = -2 and x = 3 are both asymptotes, both removable discontinuites, or a combination of the two is all dependant on the unknown numerator g(x).
(1 vote)
• Can we consider rational function as a quotient of two functions ? somewhat draw their graphs through the intersection of the functions in the numerator and the denominator ?
• A graph that is a quotient of two functions is slightly different than just a function, because a quotient of two functions creates a removable discontinuity. For example, the lines y=x and y=x²/x are the exact same, except at the x-value of 0. It results in 0 for the first function but it is undefined in the second function
• Around , Sal mentions "removable discontinuity." I watched that part 5 times but I still don't get it. Can someone explain using the example Sal provided?
• Removable discontinuities are defined in the prior section of videos. Here's a link: https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions#discontinuities-of-rational-functions

Sal's function has discontinuities at x=3 and x= - 2.
If the numerator had been defined and you we able to cancel out one of the factors in the denominator, then it would remove that discontinuity.
For example: f(x) = (x-3)(x+1) / [(x-3)(x+2)]
The factor (x-3) could be cancelled out. This would remove x=3 as a discontinuity.
Hope this helps.
• I've never come across "removable discontinuities" before, but I think I grasp the basic concept. However, why is there one only at (x-3)(...)/(x-3)(x+2), x cannot equal 3, and not one at (x+2)(...)/(x+2)(x-3), x cannot equal -2? Could someone please explain this concept to me better, or direct me to useful material? Thanks!!
• why the removable discontinuity is the specific x that makes both the denominator and numerator equal to zero?
• It's a discontinuity because plugging that value in doesn't give a number, it gives 0/0. It's a removable discontinuity because, at any point around there, whatever will make the numerator equal to 0 will cancel with whatever makes the denominator 0, and so we don't get asymptotic behavior or anything else weird.
• Why f(x) = (( x^(2)-x)) / (x^(2)-1) function has a vertical asymptote at x = -1 but not at x = 1?
• x=1 is a removable discontinuity.
If you factor the polynomials, you can see that the factor of (x-1) can be divided out which is why it is a removable discontinuity.
• So... the numerator can't be zero? Why is that?
(1 vote)
• I'm assuming you meant why the denominator can't be zero – there is no reason why the numerator can't be zero ...

If so, this is because dividing by zero is incompatible with real numbers.

Ask yourself, what is the value of 1/0?

Let's assume that there is an answer and say it equals `a`:
`1/0=a`

Take the reciprocal of both sides:
`0/1=1/a`

Multiply both sides by a:
`a*(0/1)=a*(1/a)`

But since anything times zero is still zero:
`0 = 1`

This obviously can't be true! Therefore our original assumption that `1/0=a` must be invalid ...