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I have a function whose graph is plotted below

enter image description here

First, I want to find one possible definition for the function. I know I have vertical asymptotes at $x \in \{-500, 500, 1500\}$ so $(x+500)(x-500)(x-1500)$ is in my denominator. This is of degree three.

I know that the horizontal asymptote has to be the same degree as the denominator since it is a straight line above the $x$ axis but I don't know what number it actually is. Is there a way for me to approximate aside from looking at the graph?

Picking an arbitrary number $20$, so far I have

$$f(x) = \frac{20x^3}{(x+500)(x-500)(x-1500)}$$

How do I know how many times the graph crosses the horizontal asymptote?

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  • $\begingroup$ The coefficient of $x^3$ will be the value of the horizontal asymptote. $\endgroup$
    – Landuros
    Jun 1, 2019 at 6:45
  • $\begingroup$ yes I know that but I don't know how to find that just by looking at the graph $\endgroup$
    – user477465
    Jun 1, 2019 at 6:46

1 Answer 1

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For the denominator, you're right on the mark.

The graph crosses the $x$-axis at $-2016, 1001$ and $2019$. That means the numerator must be some multiple of $(x+2016)(x-1001)(x-2019)$. The leading coefficient is equal to the horizontal asymptote. We have to guess its height, but considering the $20\,000$ tick marks on the $y$-axis, and just wildly guessing, I'd make an initial guess of about $1500$. However, I think there is a better way.

We evaluate what we have so far (with leading coefficient $1$) at $x = 0$ and get $$ \frac{2016\cdot (-1001)\cdot (-2019)}{500\cdot (-500)\cdot (-1500)} \approx 10.86 $$ From the graph, we can see that it crosses the $y$-axis just about halfway up to $20\,000$. Meaning we want the above result to be almost exactly $1000$ times larger.

Thus we land on $$ f(x) = \frac{1000(x+2016)(x-1001)(x-2019)}{(x+500)(x-500)(x-1500)} $$

How do I know how many times the graph crosses the horizontal asymptote?

From the graph, it seems obvious to me that the answer to that is $1$. But you ought to double check that the local minimum between $-500$ and $500$ is high enough, just to be sure. It crosses exactly once between $500$ and $1500$ (maybe one should double check that it's actually strictly increasing here, just to be sure), and to the left of $-500$ and to the right of $1500$ it can't cross the asymptote as it's monotonic.

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  • $\begingroup$ thank you! for the second part, I'm still unsure of determining how many times a graph crosses its horizontal asymptote.if I have the equation $\frac{1000(x+2016)(x-1001)(x-2019)}{(x+500)(x-500)(x-1500)}$ if I set it equal to the horizontal asymptote I have $\frac{1000(x+2016)(x-1001)(x-2019)}{(x+500)(x-500)(x-1500)} = 1000$ all i do is divide out the 1000 and cross multiply, and I'm left with $(x+2016)(x-1001)(x-2019)=(x+500)(x-500)(x-1500)$ From here how would i be able to tell that it crosses? $\endgroup$
    – user477465
    Jun 1, 2019 at 11:17
  • $\begingroup$ i know you said it's obvious, but I just want to know what a method could be used to check how many times it crosses it's horizontal asymptote $\endgroup$
    – user477465
    Jun 1, 2019 at 11:18
  • $\begingroup$ I said it seemed obvious from the graph. However, graphs are not exact. That equation can just be solved directly. Expand all the brackets, tidy up, and solve. And when you do, you will see that I was wrong. It crosses the asymptote once more at $x\approx 6550$. $\endgroup$
    – Arthur
    Jun 1, 2019 at 11:31
  • $\begingroup$ thanks again, is there another way of doing it instead of expanding everything out? because this was on a test of mine and I don't think that the teacher wanted us to actually expand it out $\endgroup$
    – user477465
    Jun 1, 2019 at 11:32
  • $\begingroup$ meaning, is there a way to solve $(x+2016)(x-1001)(x-2019)=(x+500)(x-500)(x-1000)$ without actually expanding everything out $\endgroup$
    – user477465
    Jun 1, 2019 at 11:32

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