This is a sneaky question,
f(x) = (x^2+3x+2)/(x+2)*tan(π*x)
= (x+1)(x+2)/(x+2) tan(πx)
= (x+1)tan(πx)
the vertical asymptote is caused by the tan(πx) part,
recall that tan90° is undefined, and 90° = π/2 radians or (1/2)π
so if x = 1/2 we have a vertical asymptote
when x = -2, we have a "hole" in our graph, which of course would not be visible in a physical graph
http://www.wolframalpha.com/input/?i=y+%3D+%28x%5E2%2B3x%2B2%29%2F%28x%2B2%29*tan%28%CF%80*x%29
Which of the following is an asymptote of f(x)=(x^2+3x+2)/(x+2)*tan(pi*x)?
A) x=-2
B) x=-1
C) x=1/2
D) x=1
E) x=2
Answer is C but I got A because substituting x for -2 makes the graph undefined.
1 answer