How do you identify all asymptotes or holes for #y=(4x^3+32)/(x+2)#?

Question

Chloe Alexander · Accepted Answer

The function is a parabola

Let's do the long division #color(white)(aaaa)##4x^3##color(white)(aaaaaaaaaaa)##+32##∣##x+2# #color(white)(aaaa)##4x^3+8x^2##color(white)(aaaaaaaaa)##∣##4x^2-8x+16# #color(white)(aaaaaa)##0-8x^2# #color(white)(aaaaaaaa)##-8x^2-16x# #color(white)(aaaaaaaa)##-8x^2-16x# #color(white)(aaaaaaaaaaaa)##0+16x+32# #color(white)(aaaaaaaaaaaaaa)##+16x+32# #color(white)(aaaaaaaaaaaaaaaa)##+0+0# The result is #(4x^3+32)/(x+2)=4x^2-8x+16# #=4(x^2-2x+4)=4(x-2)^2# This is a parabola graph{4(x-2)^2 [-25.65, 25.67, -12.83, 12.84]}

Madelyn Anderson · Answer

undefined To identify all asymptotes or holes for the function $ y = \frac{4x^3 + 32}{x + 2} $, we first need to analyze its behavior near any potential vertical asymptotes or holes. The function will have a vertical asymptote at any value of $ x $ where the denominator equals zero but the numerator does not. To find the vertical asymptotes, we solve $ x + 2 = 0 $ which yields $ x = -2 $.

To determine if $ x = -2 $ corresponds to a hole or an asymptote, we factor the numerator. Factoring $ 4x^3 + 32 $ yields $ 4(x^3 + 8) $, and then factoring further using the sum of cubes formula, $ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $, we get $ 4(x + 2)(x^2 - 2x + 4) $. Since the factor $ (x + 2) $ appears in both the numerator and the denominator, we have a hole at $ x = -2 $.

To find the $ y $-coordinate of the hole, we evaluate the function at $ x = -2 $ by plugging it into the simplified function:

$ y = \frac{4(-2)^3 + 32}{-2 + 2} = \frac{4(-8) + 32}{0} = \frac{-32 + 32}{0} = \frac{0}{0} $

Since we have an indeterminate form $ \frac{0}{0} $, we can use L'Hôpital's Rule to find the limit as $ x $ approaches $ -2 $. Differentiating the numerator and the denominator, we get:

$ y = \lim_{x \to -2} \frac{12x^2}{1} = \frac{12(-2)^2}{1} = \frac{48}{1} = 48 $

Therefore, the hole is at $ (-2, 48) $.

To summarize:
- There is a vertical asymptote at $ x = -2 $.
- There is a hole at $ (-2, 48) $.