How do you graph y=-3/(x+2)y=3x+2 using asymptotes, intercepts, end behavior?

1 Answer
Dec 17, 2016

Horizontal asymptote: y = 0. Vertical asymptote : x=-2x=2.
y-intercept: -3/2

Explanation:

(y-m_1x-c_1)(y-m_2x-c_2)= 0(ym1xc1)(ym2xc2)=0 represents the pair of lines

y = m_1x+c_1 and y=m_2x+c_2y=m1x+c1andy=m2x+c2.

(y-m_1x-c_1)(y-m_2x-c_2)= k(ym1xc1)(ym2xc2)=k represents the hyperbola having the

pair of asymptotes

(y-m_1x-c_1)(y-m_2x-c_2)= 0(ym1xc1)(ym2xc2)=0

Here, it is

y(x+2)=-3y(x+2)=3

So, the asymptotes are given by

y = 0 and x+2=0.

graph{y(x+2)+3=0x^2 [-40, 40, -20, 20]}