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

Question

Caleb Alexander · Accepted Answer

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

#(y-m_1x-c_1)(y-m_2x-c_2)= 0# represents the pair of lines #y = m_1x+c_1 and y=m_2x+c_2#. #(y-m_1x-c_1)(y-m_2x-c_2)= k# represents the hyperbola having the pair of asymptotes #(y-m_1x-c_1)(y-m_2x-c_2)= 0# Here, it is #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]}

James Allen · Answer

undefined To graph $ y = -\frac{3}{{x + 2}} $:

1. **Vertical Asymptote**: Set the denominator equal to zero and solve for $ x $. Here, $ x + 2 = 0 $, so $ x = -2 $ is the vertical asymptote.

2. **Horizontal Asymptote**: When $ x $ approaches positive or negative infinity, $ y $ approaches a certain value. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $ y = 0 $.

3. **Intercepts**: To find the $ y $-intercept, set $ x = 0 $ and solve for $ y $. $ y = -\frac{3}{{0 + 2}} = -\frac{3}{2} $. So, the $ y $-intercept is at $ (0, -\frac{3}{2}) $. To find the $ x $-intercept, set $ y = 0 $ and solve for $ x $. $ 0 = -\frac{3}{{x + 2}} $ has no real solution, so there's no $ x $-intercept.

4. **End Behavior**: As $ x $ approaches negative infinity, $ y $ approaches 0 from below. As $ x $ approaches positive infinity, $ y $ approaches 0 from above.

Given this information, the graph has a vertical asymptote at $ x = -2 $, a horizontal asymptote at $ y = 0 $, a $ y $-intercept at $ (0, -\frac{3}{2}) $, and the end behavior mentioned above.