Answer to Question #92210 in Algebra for Karla Carreras

Question

Identify the vertical and horizontal asymptotes for (2x+4)/(x^2+5x+6)
** please answer by tommorow!!

Answer 1

$$f(x)=\frac{2x+4}{x^2+5x+6}=\frac{(x+2)}{(x+2)(x+3)}=\frac{2}{x+3}$$

It has the discontinuities at x = -3 and at x = -2 (also a puncture point)

1) x = -3 is a vertical asymptote because

$$\lim_{x\to-3}f(x)=\lim_{x\to-3}\frac{2}{x+3}=\infty$$

2) x = -2 is not a vertical asymptote because

$$\lim_{x\to-2}f(x)=\lim_{x\to-2}\frac{2}{x+3}=2$$

Let the slant (or a horizontal asymptote in particular) asymptote be

$$y=kx+b$$

$$k=\lim_{x\to\infty}(\frac{f(x)}{x})=\lim_{x\to\infty}(\frac{2}{x^2+3x})=\lim_{x\to\infty}(\frac{\frac{2}{x^2}}{1+\frac{3}{x}})=0$$

$$b=\lim_{x\to\infty}(f(x)-kx)=\lim_{x\to\infty}(\frac{2}{x+3})=\lim_{x\to\infty}(\frac{\frac{2}{x}}{(1+\frac{3}{x})})=0$$

Thus, the function has a horizontal asymptote y = 0 and the vertical asymptote x = -3.