Question

find the horizontal and vertical asymptotes for the function. f(x) = (7x³ + x² - 3)/(x³ + 8) vertical asymptote: x = ? horizontal asymptote: y =

Explanation:

Step1: Find Vertical Asymptote

To find the vertical asymptote, we set the denominator equal to zero and solve for \(x\). The denominator is \(x^3 + 8\). We can factor this using the sum of cubes formula \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\), where \(a = x\) and \(b = 2\) (since \(2^3=8\)). So, \(x^3+8=(x + 2)(x^2-2x + 4)\). Setting the denominator equal to zero:
\(x^3 + 8 = 0\)
\((x + 2)(x^2-2x + 4)=0\)
The quadratic factor \(x^2-2x + 4\) has discriminant \(\Delta=(-2)^2-4\times1\times4=4 - 16=-12<0\), so it has no real roots. The real root comes from \(x + 2 = 0\), so \(x=-2\). We also need to check that the numerator is not zero at \(x = - 2\). Substitute \(x=-2\) into the numerator: \(7(-2)^3+(-2)^2-3=7(-8)+4 - 3=-56 + 1=-55
eq0\). So the vertical asymptote is at \(x=-2\).

Step2: Find Horizontal Asymptote

For rational functions, if the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. The degree of the numerator \(7x^3+x^2 - 3\) is 3, and the degree of the denominator \(x^3 + 8\) is 3. The leading coefficient of the numerator is 7, and the leading coefficient of the denominator is 1. So the horizontal asymptote is \(y=\frac{7}{1}=7\).

Answer:

Vertical asymptote: \(x = -2\)
Horizontal asymptote: \(y = 7\)