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## Precalculus

### Course: Precalculus > Unit 4

Lesson 4: Graphs of rational functions- Graphing rational functions according to asymptotes
- Graphs of rational functions: y-intercept
- Graphs of rational functions: horizontal asymptote
- Graphs of rational functions: vertical asymptotes
- Graphs of rational functions: zeros
- Graphs of rational functions

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# Graphs of rational functions: y-intercept

Sal picks the graph that matches f(x)=(ax^m+bx+12)/(cx^m+dx+12) based on its y-intercept.

## Want to join the conversation?

- Choice C actually has a y-intercept of -2 not -1(29 votes)
- What is a removable discontinuity?(3 votes)
- It's when you have a function, and there is a "hole" in the function at a certain x-value. If you placed just 1 point on that gap, the function would be normal - hence the name
*removable*discontinuity. For example, go to some graphing system and input y=x^3/x. The function is undefined at x=0, but with no odd behavior near it.(5 votes)

- What does the U shaped figure inside the Asymptotes mean? Some or U pointing up and some are U shaped pointing down.

Thanks in advance.(2 votes)- Those are parts of the function curves.

For example in the "A" graph, at the left-most asymptote the function changes sign and the curve jumps from −∞ to +∞.(2 votes)

- Rational functions can have 3 separate parabolas ?(2 votes)
- Technically they aren't parabolas because they have asymptotes.(2 votes)

- is there an easier video about asymptotes? I'm in 11th grade and have never learned this before, sad:((2 votes)
- An asymptote is when the denominator equals zero. You can get this from a equation by factoring. Different from a removable discontinuity which is the other number you get when you factor but divide from the numerator as well.(2 votes)

- In the first line, the word 'constants', is mentioned. In the rational expression, only 12 is a constant, so why does it say the following; 'a, b, c, and d are unknown constants.'?(2 votes)
- Coefficients are constant values. "a, b, c and d" are the coefficients.(2 votes)

- In the rational expression, we have 'ax^n' and 'cx^m'. Why can't they both be ^n or ^m?(2 votes)
- This is likely to ensure that the 2 terms are unlike. If the exponents matched, you could add the two terms as they would be like terms.(2 votes)

- What's the difference between integers and unknown constants?

When it says towards the beginning; 'where m and n are integers.....'

Why couldn't it just be written as integers for both, or unknown constants for both. I don't understand why they've used integers for exponents and unknown constants for the co-efficients?(2 votes)- Integers would not involve fractions or decimals. So, the exponents are being restricted to negative whole numbers, 0, or positive whole number. I think later in the video a correction box tells you to restrict the exponents to positive integers. This is likely done to keep the numerator and denominator as polynomials. Polynomials require positive exponents on the variables.

The coefficients are not being restricted - they can be any real number.(2 votes)

- in which cases do rational functions have three parabolas, please?(2 votes)
- At1:22, x can be zero and the whole thing can be zero. But, if n and m are 0 then that term would be 1. How is this possible? Or if m is zero and n is something else, it would be 12/13.(0 votes)
- You are right, in the m = 0 or n = 0 case, the intercept could be 12/13 or 13/12 instead of 1.

Of course, none of the graphs have a y-intercept of 12/13 or 13/12 so B remains the only possible choice.(4 votes)

## Video transcript

- [Voiceover] Let f of
x equal a times x to n plus b x plus 12 over c times x to the m plus d x plus 12 where m and n are integers and a, b, c and d are
unknown constants are. This is interesting. Which of the following is a possible graph of y is equal to f of x? And it tells us the dashed
lines indicate asymptotes. So they gave us four choices here. And so I encourage you like always. Pause this video. Give a go at it. See if you can figure
out which of these graphs could be the graph for y equals f of x where f of x, where they've given us this information about f of x. All right, now let's
work to this together. Now this is really interesting. It's, you know, they didn't give us a lot. They didn't even tell us
what the exponents on x are. They haven't even given
us the coefficients. All they've told us is this 12 here. So this 12 looks like a pretty big clue. So what can that tell us? The way they've written that we really can't deduce any
zeros for the function. We really can't deduce what x values make the numerator equals zero or what x values make the
denominator equal zero. So it's gonna be hard for us to deduce what are the zeros of the function or what are the removable discontinuities or what are the vertical asymptotes. But just these 12 sitting
here does tell us one thing. What happens when x equals zero? Because when x equal
zero, every other term in this rational expression is just going to be equal to zero. And so we can figure out f of zero. F of zero is going to be equal to, well a times zero to the n, well that's just going to be a zero plus b times zero. Well, that's just going to be zero. Plus 12 over c times zero to the m power. Well, that's just going to be zero. D times zero is going to be a zero. And then we have our 12 there. So we're actually able to figure out what f of zero is. It's 12 over 12 or one. So we actually know the y
intercept for this function. Let's see if that's enough information for us to figure out
if any of these choices could be the graph of y equals f of x. So let's see here. So choice a, our y intercept is a two. When x equals zero, our graph goes to two. So we can rule that out. The y intercept needs to be one. So we can rule out, let's see, choice B does have a y intercept. It looks just eyeballing it at one, x equals zero, y is one. So this looks interesting. Choice C has its y intercept at y is equal to negative one. So once again, we can rule that out. And choice D has no y-intercept at all. So that was enough information. And lucky for us because
they really didn't give us a lot more information than just what be able to evaluate f of zero. We were able to figure
out any of the other, you know, the zeros or
the vertical asymptotes or their removable discontinuities. So we definitely feel
good about that choice.