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

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Answer 1 · 2016-12-17T16:01:34

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

$(y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = 0$ $(y-m_1x-c_1)(y-m_2x-c_2)= 0$ represents the pair of lines

$y = m_{1} x + c_{1} and y = m_{2} x + c_{2}$ $y = m_1x+c_1 and y=m_2x+c_2$ .

$(y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = k$ $(y-m_1x-c_1)(y-m_2x-c_2)= k$ represents the hyperbola having the

$(y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = 0$ $(y-m_1x-c_1)(y-m_2x-c_2)= 0$

Here, it is

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

So, the asymptotes are given by

y = 0 and x+2=0.

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

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