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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: horizontal asymptote

Sal picks the graph that matches f(x)=(-x²+ax+b)/(x²+cx+d) based on its horizontal asymptote.

## Want to join the conversation?

- Is there a way that I can understand the logic behind this proof. Technically it makes sense, but logically it doesn't add up to me. Thanks in advance!(18 votes)
- The question seeks to gauge your understanding of horizontal asymptotes of rational functions. The behavior of rational functions (ratios of polynomial functions) for large absolute values of x (Sal wrote as x goes to positive or negative infinity) is determined by the highest degree terms of the polynomials in the numerator and the denominator. This particular function has polynomials of degree 2 in both the numerator and the denominator. Thus for values of x far away from zero we can approximate the ratio by considering only the dominant (degree 2) terms. that is why Sal simplified the ratio to -x^2/x^2 which is just -1. It is not "equal" to -1 for large values of but it is "close".

For example, Suppose x is 1,000,000. Then x^2 will be 10^12 and -x^2 will be -10^12. The other terms in the numerator and denominator will all be hundreds of thousands of times smaller than a trillion. So -1trillion-ish/1 trillion-ish is pretty close to -1. This assumes that our unknown constants aren't HUGE, but if they are, pick an x big enough to make them look "small".(21 votes)

- What if the powers of the highest degree terms in the denominator and numerator are not the same?(9 votes)
- if the numerator degree is lower than the denominator degree, then the horizontal asymptote is y=0 because as x gets really large or small the denominator will be much larger than the numerator, so y will be around 0.

if the numerator degree is higher than the denominator degree, then divide the whole numerator expression by the whole denominator. ignore the remainder you get because that will approach 0. the asymptote is the part of the quotient that isn't the remainder. the asymptote can be linear, quadratic, cubic or any degree polynomial(10 votes)

- I have a tricky problem: When you have a function:

y= 1/(5/x-3) so you have a fraction in a fraction: "1 over 5 over x-3", then something strange appears to me :

if you would set x=3 you will receive an error (you would try to divide by zero), but if you transform the equation to : y = 1 * (x-3/5) you are allowed to use x=3.

Also if you put y= 1/(5/x-3) in ANY graphic calculator the calculator will map you a graph that shows x=3 is a valid point on the graph.**Question: How can it be that two versions of the same equation allow you different values?**

Put x= 3 into y=1 / (5/x-3) and you receive an error, transform it to y=1 * (x-3/5) and you receive a total valid value plus a perfect graph. Why is this possible?

bold

The same thing happens for the graph of cotan (x).

Technically cot(x)= 1/ (sin(x)/cos(x)) = 1 over (sin x over cos x) or cot(x)= 1 / tan(x). Whenever cos(x)=0 the equation cot(x)= 1/ (sin(x)/cos(x)) is going to be invalid, but the graph of cotan maps those values for which cos(x)= 0 and for which tan (x) would be invalid .**Question (other way) : How is it possible to have a value for x (like 90°) for which tan(x)=undefined giving a perfect mapped graph in 1/tan(x) ? Is it not the same like y= 1 / Undefined?**

*bold

Hope you see my problem.(8 votes)- 𝑓(𝑥) = 1/[5/(𝑥 – 3)] is not the same function as 𝑔(𝑥) = (𝑥 – 3)/5. This is because the domain of 𝑓 is different from the domain of 𝑔. At every point except for 𝑥 = 3, the function 𝑓 behaves exactly the same as 𝑔 but because of that one discontinuity, they are
*not the same function*. Also, my graphing calculator graphed 𝑓 with a hole at 𝑥 = 3 so you might not be looking close enough or you might have input the function incorrectly. It just so happens that the discontinuity occurs at (3, 0) which is on the 𝑥-axis which could explain why you couldn't see an apparent "hole". If you are using a TI calculator, you can open the table of input/outputs and you'll notice that at 𝑥 = 3, the table returns an error for 𝑦 as desired.

Also:

cot 𝑥 ≠ 1/[(sin 𝑥)/(cos 𝑥)]

Instead, cotangent is defined as the completely simplified version of the expression above. This is (cos 𝑥)/(sin 𝑥). In other words:

𝑓(𝑥) = (cos 𝑥)/(sin 𝑥) = cot 𝑥

and:

𝑔(𝑥) = 1/[(sin 𝑥)/(cos 𝑥)]

Then 𝑓 and 𝑔 are not identical functions! The idea of domain restrictions before simplifying a rational expressions is important, and though the function after simplification will behave very similar to the function before, they are*not the same function*! This is also important when it comes to finding the domains of compositions of rational functions. Comment if you have questions!(9 votes)

- How do you find the horizontal asymptote if the degree is not the same in both the dividend and the divisor? Is there another video about that?(7 votes)
- "When the degree of the numerator of a rational function is less than the degree of the denominator, the x-axis, or y=0, is the horizontal asymptote.

When the degree of the numerator of a rational function is greater than the degree of the denominator, there is no horizontal asymptote."

—*College Algebra: Graphs and Models*by Marvin Bittinger, et al.(5 votes)

- If I have something unusual, like :

Would my asymptote be 2 or 3/4?`6x^6 - 3x^4 + 6x`

----------------

3x^6 + 4x^4(5 votes)- Your horizontal asymptote would y = 2. If the degree of the numerator and denominator are the same, then you have to divide the leading co-efficients.(6 votes)

- When does he explain oblique asymtotes and how to use "long division" to find them?(5 votes)
- At the1:35mark, it was determined that f(x) = -1. But in the next comment, why does Sal then say... "So now we have "y" equal to -1 (as opposed to "x" = -1?(3 votes)
- f(x) is Y

They are interchangeable. If you have an equation y=x+5, it is written in function notation as f(x)=x+5.

Hope thi shelps.(5 votes)

- is not -x^2 always the same as x^2 (because a negative number times itself is always positive), which would make the horizontal asymptote of this function = 1/1 = 1, not -1.

(Unless maybe the question was written with -(x^2) )

Thanks for clarifying - I love KA!(3 votes)- The difference between -x^2 and (-x)^2 is this: the first one will always have a negative answer, because even though x^2 is always positive, that negative sign will make it negative. The second one will be positive, because a negative number, in this case -x, raised to the power of 2 is always positive.

For example, -2^2 is -4. If you put it in your calculator like that, with no parenthesis, that's what you'll get. But if you enter (-2)^2, the answer will be 4.

I hope this is what you were asking and that this helps clarify things for you!(4 votes)

- what about when the numerator has a higher power of x than the denominator? like for example= f(x)=x^2+3/x+2(3 votes)
- As the degree in the numerator is higher than the degree in the denominator, there will be no horizontal asymptote.

The general rule of horizontal asymptotes, where n and m is the degree of the numerator and denominator respectively:

n < m: x = 0

n = m: Take the coefficients of the highest degree and divide by them

n > m: No horizontal asymptote :)(2 votes)

- what is the derivative of s(x)=4e^x-3(1 vote)
- The derivative of s(x) = 4e^x - 3 with respect to x is:

s'(x) = d/dx (4e^x - 3)

= d/dx (4e^x) - d/dx (3) [since the derivative of a constant is zero]

= 4d/dx (e^x) - 0 [since the derivative of e^x is e^x]

= 4e^x

Therefore, s'(x) = 4e^x.(1 vote)

## Video transcript

- [Voiceover] Let f of x
equal negative x squared plus a x plus b over x squared plus c x plus d where a, b, c, and d
are unknown constants. Which of the following is a possible graph of y is equal to f of x? And they told us dashed
lines indicate asymptotes. So this is really interesting here. And they gave us four choices. We see four of them,
three of them right now. Then if I scroll over bit over, you can see choice D. And so I encourage you to pause the video and think about how we can figure it out because it is interesting because they haven't
given us a lot of details. They haven't given us
what these coefficients or these constants are going to be. All right, now let's think about it. So one thing we could think about is horizontal asymptotes. So let's think about what happens as x approaches positive
or negative infinity. Well, as x, so as x approaches infinity or x approaches negative infinity, f of x. F of x is going to be
approximately equal to. Well, we're going to look
at the highest degree terms because these are going to dominate as the magnitude of x, the absolute value of
x becomes very large. So f of x is going to be
approximately negative x squared over x squared which is equal to negative or we could another way to think about it. This is the same thing as negative one. So f of x is going to approach, f of x is going to approach negative one. In either direction, as
x approaches infinity or x approaches negative infinity. So we have a horizontal asymptote at y equals negative one. Let's see, choice A here, it does look like they
have a horizontal asymptote at y is equal to negative one right over there. And we can verify that because each hash mark is two. We go from two to zero to negative two to negative four. So this does look like
it's a negative one. So just based only on
the horizontal asymptote, choice A looks good. Choice B, we have a horizontal asymptote at y is equal to positive two. So we can rule that out. We know that a horizontal asymptote as x approaches positive
or negative infinity is at negative one, y equals negative one. Here, our horizontal asymptote is at y is equal to zero. The graph approaches, it approaches the x axis from either above or below. So it's not the horizontal asymptote. It's not y equals negative one. So we can rule that one out. And then similarly, over here, our horizontal asymptote is
not y equals negative one. Our horizontal asymptote
is y is equal to zero so we can rule that one out. And that makes sense because really they only
gave us enough information to figure out the horizontal asymptote. They didn't give us enough information to figure out how many roots or what happens in the interval and all of those types of things. How many zeros and all that because we don't know what the actual coefficients or constants
of the quadratic are. All we know is what happens as
the x squared terms dominate. This thing is going to
approach negative one. And so we picked choice A.