Respuesta :
The oblique asymptote in the slope-intercept form is:
y = m x + b, where:
m = [tex] \lim_{x \to \infty} \frac{ 2x^{2}+3x+8}{ x^{2} +3x} [/tex] = 2
b = [tex] \lim_{x \to \infty} \frac{ x^{2} +3x+8}{x+3} - 2 x = \\ \lim_{x \to \infty} \frac{2 x^{2} + 3 x + 8 - 2 x^{2} -6x}{x+3}= \\ \lim_{x \to \infty} \frac{-3x+8}{x+3} [/tex] = -3
Answer: y = 2 x - 3
y = m x + b, where:
m = [tex] \lim_{x \to \infty} \frac{ 2x^{2}+3x+8}{ x^{2} +3x} [/tex] = 2
b = [tex] \lim_{x \to \infty} \frac{ x^{2} +3x+8}{x+3} - 2 x = \\ \lim_{x \to \infty} \frac{2 x^{2} + 3 x + 8 - 2 x^{2} -6x}{x+3}= \\ \lim_{x \to \infty} \frac{-3x+8}{x+3} [/tex] = -3
Answer: y = 2 x - 3