How do you identify all asymptotes or holes for $y = (2 x + 1) - \frac{3}{x - 4}$ $y=(2x+1)-3/(x-4)$ ?

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Answer 1 · 2016-11-01T08:48:55

There are 2 asymptotes,
a vertical one $x = 4$ $x=4$
and a slant one $y = 2 x + 1$ $y=2x+1$

A vertical asymptote is $x = 4$ $x=4$ because we cannot divide by $0$ $0$
The slant asymptote is $y = 2 x + 1$ $y=2x+1$

graph{(y-2x-1-3/(x-4))(y-2x-1)=0 [-14.24, 14.23, -7.12, 7.12]}

How do you identify all asymptotes or holes for y=(2x+1)-3/(x-4)y=(2x+1)−3x−4y=(2x+1)-3/(x-4)?