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let’s say we have a function $$ f(x)=\frac{2x^2-5x+2}{x^2-4} $$ finding the ratio of the leading terms gives you $2x^2/x^2=2$, so we have a horizontal asymptote at $y = 2$. If you factor out the denominator and numerator, you get $$ f(x) = \frac{(2x-1)(x-2)}{(x-2)(x+2)} $$ Vertical asymptote at $x = -2$, hole at $x=2$.

Simplifying gives you $(2x-1)/(x+2)$. If we want to find the intersection of a function to the horizontal asymptote, we set it equal to the asymptote $y = 2$.

If you use the main function: $$ \frac{2x^2-5x+2}{x^2-4}=2 \Longrightarrow 2x^2-5x+2=2(x^2-4)\Longrightarrow 2x^2-5x+2 =2x^2-8, $$ simplifying and solving for x gives $x=2$. So an intersection at $x=2$.

If you use the simplified version of the function: $$ \frac{2x-1}{x+2} = 2\Longrightarrow 2x-1=2(x+2)\Longrightarrow 2x-1=2x+4\Longrightarrow -1=4, $$ that is an invalid statement, meaning that you do not get an intersection.

Which one is correct???

eyeballfrog
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Abdulmalek Alem
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1 Answers1

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The problem is here:

$$ \frac{2x^2-5x+2}{x^2-4}=2 \Longrightarrow 2x^2-5x+2=2(x^2-4) $$

You multiplied both sides by $x^2 - 4$, and then found $x = 2$. But that means $x^2 - 4$ was $0$, and multiplying both sides of an equation by $0$ always gives you the true statement $0 = 0$. So not helpful.

You should use the simplified version to check for intersections. If an intersection exists in the function's domain, the simplified version will find it. However, it may also find values outside the function's domain (for example a horizontal asymptote passing through a removable singularity). So you need to check any answers you find against that.

eyeballfrog
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