Given \(y = \frac{2}{f(x)}\text{, 2 is actually an }a\text{ value}\) (mapping rule \((af(k(x - p)) + q\to(\frac{1}{k}x + p, ay + q)\))
Given a linear reciprocal \(\frac{1}{kx - c}\) the restriction of not dividing by zero will give \(x\neq\frac{c}{k}\) as a restriction. \(x = \frac{c}{k}\) also becomes the vertical asymptote.
When testing vertical asymptotes, plug in numbers that are slightly larger than the value of the asymptote from both sides. Eg. for \(\frac{1}{x - 3}\) plug in 2.99 and 3.01 to see if the number is positive or negative. If the number is negative that side of the asmyptote shoots down. If the number is positive, that side of the asymptote shoots up.
From the negative side (plugging in 2.99 because \(2.99 < 3\) therefore is \(3^-\)) the function shoots down because plugging in a number from the negative side returns a negative value
From the positive side (plugging in 3.01 because \(3.01 > 3\) therefore is ##3^+$$) the function shoots up because plugging in a number from the positive side returns a positive
To test behaviors as the function approaches \(\pm\infty\) plug in super huge numbers like \(\pm 999\)
Plugging in -999 into the equation returns a number less than 0 (the horizontal asymptote is \(y = 0\)) and plugging in 999 returns a number larger than 0 (horizontal asymptote). Therefore when \(x\to -\infty\) it is below the horizontal asymptote and when \(x\to +\infty\) it is above the horizontal asymptote.