Question

identify the asymptotes. give your answers in exact form. do not round.

q(x)=\frac{x^{3}+4 x^{2}-3 x+1}{x^{2}-5}

separate multiple equations of asymptotes with commas as necessary. select
one\ if applicable.

part 1 of 2

equation(s) of the vertical asymptote(s):

part 2 of 2

equation(s) of the horizontal asymptote(s):

equation(s) of the slant asymptote(s):

Explanation:

Step1: Find vertical asymptotes

Set the denominator \(x^{2}-5 = 0\).
Using the formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a=x\) and \(b=\sqrt{5}\), we have \((x+\sqrt{5})(x - \sqrt{5})=0\).
Solving \(x+\sqrt{5}=0\) gives \(x=-\sqrt{5}\), and solving \(x-\sqrt{5}=0\) gives \(x=\sqrt{5}\).

Step2: Check for horizontal asymptote

The degree of the numerator \(n = 3\) (since the highest - power term in \(x^{3}+4x^{2}-3x + 1\) is \(x^{3}\)) and the degree of the denominator \(m=2\) (since the highest - power term in \(x^{2}-5\) is \(x^{2}\)).
When \(n>m\) (here \(n - m=3 - 2 = 1\)), there is no horizontal asymptote.

Step3: Find slant asymptote

Use polynomial long - division.
Divide \(x^{3}+4x^{2}-3x + 1\) by \(x^{2}-5\).

$$
LATEXBLOCK0
$$

As \(x\to\pm\infty\), \(\frac{2x + 21}{x^{2}-5}\to0\). So the slant asymptote is \(y=x + 4\).

Answer:

Part 1 of 2

Equation(s) of the vertical asymptote(s): \(x = \sqrt{5},x=-\sqrt{5}\)

Part 2 of 2

Equation(s) of the horizontal asymptote(s): None
Equation(s) of the slant asymptote(s): \(y=x + 4\)