I have a graph with the equation: $f(x)=\frac{1000(x+2016)(x-1001)(x-2019)}{(x+500)(x-500)(x-1000)}$
and it looks like this:
I want to know a way of determining how many times (if at all) this graph crosses its horizontal asymptote. I found a video where someone said to set the equation equal to its horizontal asymptote and solve. So
$\frac{1000(x+2016)(x-1001)(x-2019)}{(x+500)(x-500)(x-1000)}=1000$
Dividing by 1000 and cross multiplying, I get:
$(x+2016)(x-1001)(x-2019)=(x+500)(x-500)(x-1000)$
Without expanding all this out, how can I see how many times the graph crosses its horizontal asymptote.
I know it looks like it does on the right but I want to know how many times it crosses and where.
Best Answer
You have
$$ (x+2016)(x-1001)(x-2019) = (x+500)(x-500)(x-1000) \Leftrightarrow \\ (x+2016)(x-1001)(x-2019) - (x+500)(x-500)(x-1000)=0 $$ It shouldn't be too hard to see that the $x^3$ term cancels. The contributiors to the $x^2$ terms are $$ - 2019 \\ +2016 \\ - 1001 \\ + 1000 \\ + 500 \\ - 500 $$ Which adds up to $-4$ $\Rightarrow $ you will get a second degree polynomial.
A second degree polynomial has two roots. They can either be real or complex. So you either has $2$ times of crossing, or $0$ times. Which of them is it?
Set $f(x) = 1000 \frac{(x+2016)(x-1001)(x-2019)}{ (x+500)(x-500)(x-1000)}$ then $f(1000,05) > 1000$ and $f(3000) < 1000$. Since $f$ is continuous between $x_1 = 1000,05$ and $x_2 = 3000$, by the intermediate value theorem $f$ must cross $1000$. Using the fact that it can only cross two timers och never. There has to be two crossings.