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### Course: Precalculus>Unit 4

Lesson 4: Graphs of rational functions

# Graphs of rational functions: zeros

Sal picks the graph that matches f(x)=(2x²-18)/g(x) (where g(x) is a polynomial) based on its zeros.

## Want to join the conversation?

• The example Sal gives at ,
(2(x+3)(x-1))/ (x+3)^2(x+1) I don't understand why x at -3 isn't a removable discontinuity. If you put -3 in the original equation without simplying, you get 0/0.

Thank you.
• A removable discontinuity would divide COMPLETELY out of the denominator.
In this case, after the division there is still a factor of ( x + 3 ) left in the denominator,
so that takes precedence over the fact that a single ( x + 3 ) was able to be divided out
and indicates an asymptote at x = -3 instead of a removable discontinuity.
• So I'm still confused about how you would solve a problem like this on your own, without knowing which random examples to pick?
• g(x) can be anything, so Sal is choosing factors that cancel out factors in the numerator or create asymptotes.
• Hello all, could someone please explain how the answer choice is A? Thanks :)
• The zeros of the numerator are -3 and 3. So, at x = -3 and x = 3, the function should have either a zero or a removable discontinuity, or a vertical asymptote (depending on what the denominator is, which we do not know), but it must have either of these three “interesting” behaviours at x = -3 and x = 3. The answer A is the only one that has “interesting” behaviour at both x = -3 and x = 3.
• By the end of the video, Sal disregard the choice letter c. Isn't it possible to have a discontinuity at x = 3 since we don't exactly know what g(x) is? It could have and (x-3) or (x+3) so it's possible to have a discontinuity at 3 right?
(1 vote)
• It is certainly possible to have a discontinuity at x = 3, but the problem with choice c is that its asymptote is not at x = -3.
Listen again carefully from about to where Sal sums up various situations that may be indicated by the numerator.
• How would you know what t do even you dont understand the information?
• at why sal is pertaining to x-intercepts when he is talking about zeros of the function?
• A zero of a function is the value of x that makes the function value zero.
All points with a function value of zero lie on the X-axis.
In other words, the function intercepts the X-axis at those points.
(1 vote)
• Why can't 3 be a vertical asymptote?
• Sorry for necroposting but I wanted to clarify this for future readers.

Since g(x) can be any polynomial (no other info is given about it), it actually can if g(x) = (x - 3)². If that were the case, then f(x) would simplify to 2(x + 3) / (x - 3), which has a zero at x = -3 and a vertical asymptote at x = 3. But the choices don't include that which is fine since they only want you to pick one possibility.

The only thing you need to note is something interesting must happen at both x = -3 and x = 3. As a pair, they need to be a combination of zeros, vertical asymptotes, and/or removable discontinuities.
(1 vote)
• sal seid We've seen that situation where something
that makes the numerator equals zero
could be a removable discontinuity
if you have that same expression in the denominator.
why is that?
(1 vote)
• Consider a function f(x)=(x-1)/(x-1). The numerator will be 0 for x=1, but then you will also have 0 in the denominator, and division by zero is still an undefined operation. So the graph of this function will have a "gap", or removable discontinuity at x=1.
• why -1 is not vertical asymtote?
(1 vote)
• Because nothing about this function suggests that it is. You're not given any information about g(x), so there's no reason to assume that there has to be a vertical asymptote at this x value or any other.
What Sal written to the right are just examples meant to show that the zeros you found in the numerator could also be removable discontinuities or asymptotes.
• Using GeoGebra I generated graph C with the function, f(x) = 2(x+3)(x-3)/(x+2)(x-3), on the other hand, graph A is reversed. Although it does fits the conditions...
(1 vote)
• 𝑓(𝑥) = 2(𝑥 + 3)(𝑥 − 3)∕((𝑥 + 2)(𝑥 − 3)) has a horizontal asymptote 𝑦 = 2.
Also, 𝑓(−3) = 0 and 𝑓(0) = 3.

None of those features are present in Graph C.

– – –

Graph A has a vertical asymptote at 𝑥 = −3,
which tells us that 𝑔(𝑥) must have a factor (𝑥 + 3)².

Also, graph A has a horizontal asymptote 𝑦 = −2,
which tells us that 𝑔(𝑥) must be of degree 2 and have a leading coefficient of −1.

Thus, our only choice is 𝑔(𝑥) = −(𝑥 + 3)²
⇒ 𝑓(𝑥) = (2𝑥² − 18)∕(−(𝑥 + 3)²)

Luckily, 𝑓(3) = 0 and 𝑓(0) = 2, just like graph A.