+250 A verticle asymptote occurs where a value of x makes the denominator equal to zero but not the numerator:
In this case the only value of x that makes the denominator = 0 is x=-1/2 and this value also doesn't make the numerator = 0 therefore the vertical asymptote is at x=-1/2.
What if the function was this however: $$y(x)=\frac{(x^2+7x+10)(2x+1)}{2x+1}$$
If you dont cancel the 2x+1 and substitute in x=-1/2 you will get 0 for both the numerator and the denominator therefore this wouldnt be a vertical asymptote. The way to go about this situation is to cancel the 2x+1 to get $$y(x) = x^2+7x+10$$ but you Have to mention that x cannot equal -1/2 as this isn't defined for the original function and is a discontinuity and then you can leave it at that.
Saul from Khan Academy Explains this nicely: https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example
+250 A verticle asymptote occurs where a value of x makes the denominator equal to zero but not the numerator:
In this case the only value of x that makes the denominator = 0 is x=-1/2 and this value also doesn't make the numerator = 0 therefore the vertical asymptote is at x=-1/2.
What if the function was this however: $$y(x)=\frac{(x^2+7x+10)(2x+1)}{2x+1}$$
If you dont cancel the 2x+1 and substitute in x=-1/2 you will get 0 for both the numerator and the denominator therefore this wouldnt be a vertical asymptote. The way to go about this situation is to cancel the 2x+1 to get $$y(x) = x^2+7x+10$$ but you Have to mention that x cannot equal -1/2 as this isn't defined for the original function and is a discontinuity and then you can leave it at that.
Saul from Khan Academy Explains this nicely: https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example
