# How do you find the asymptotes for (x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)?

Jul 29, 2016

Horizontal asymptote (left and right): $y = 1$

Vertical asymptotes at $x = 1$, $x = 2$, $x = - 2$

Hole at $\left(5 , \frac{9}{7}\right)$

#### Explanation:

Let's attempt to factorise the numerator and denominator first.

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Numerator

Substituting $x = - 1$ in the numerator we get:

$1 + 3 - 21 - 43 + 60 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{4} - 3 {x}^{3} - 21 {x}^{2} + 43 x + 60 = \left(x + 1\right) \left({x}^{3} - 4 {x}^{2} - 17 x + 60\right)$

Substituting $x = 3$ in the remaining cubic, we get:

$27 - 36 - 51 + 60 = 0$

So $x = 3$ is a zero and $\left(x - 3\right)$ a factor:

${x}^{3} - 4 {x}^{2} - 17 x + 60 = \left(x - 3\right) \left({x}^{2} - x - 20\right) = \left(x - 3\right) \left(x - 5\right) \left(x + 4\right)$

In summary:

${x}^{4} - 3 {x}^{3} - 21 {x}^{2} + 43 x + 60 = \left(x + 1\right) \left(x - 3\right) \left(x - 5\right) \left(x + 4\right)$

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Denominator

Substituting $x = 1$ in the denominator we get:

$1 - 6 + 1 + 24 - 20 = 0$

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{4} - 6 {x}^{3} + {x}^{2} + 24 x - 20 = \left(x - 1\right) \left({x}^{3} - 5 {x}^{2} - 4 x + 20\right)$

The remaining cubic factors by grouping:

${x}^{3} - 5 {x}^{2} - 4 x + 20$

$= \left({x}^{3} - 5 {x}^{2}\right) - \left(4 x - 20\right)$

$= {x}^{2} \left(x - 5\right) - 4 \left(x - 5\right)$

$= \left({x}^{2} - 4\right) \left(x - 5\right)$

$= \left(x - 2\right) \left(x + 2\right) \left(x - 5\right)$

In summary:

${x}^{4} - 6 {x}^{3} + {x}^{2} + 24 x - 20 = \left(x - 1\right) \left(x - 2\right) \left(x + 2\right) \left(x - 5\right)$

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Together

$\frac{{x}^{4} - 3 {x}^{3} - 21 {x}^{2} + 43 x + 60}{{x}^{4} - 6 {x}^{3} + {x}^{2} + 24 x - 20}$

$= \frac{\left(x + 1\right) \left(x - 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}} \left(x + 4\right)}{\left(x - 1\right) \left(x - 2\right) \left(x + 2\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 5\right)}}}}$

$= \frac{\left(x + 1\right) \left(x - 3\right) \left(x + 4\right)}{\left(x - 1\right) \left(x - 2\right) \left(x + 2\right)}$

with exclusion $x \ne 5$

Looking at the leading terms of the numerator and denominator, the degrees and coefficients are identical, so there is a horizontal asymptote (left and right): $y = 1$

When $x = 1$ or $\pm 2$ the denominator is zero and the numerator non-zero, so these values of $x$ correspond to vertical asymptotes.

Substituting the value $x = 5$ into the simplified rational function, we find the coordinate of the hole as: $\left(5 , \frac{9}{7}\right)$

graph{(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20) [-41.83, 38.17, -10.24, 29.76]}