Question

You will compare the types of graphs in 11.3 with those in this lesson. Graph \(f(x)=\frac{6}{x}\) and \(f(x)=\frac{6}{x-2}+1\) in the same coordinate plane.

Short answer

The graph of \(f(x) = \frac{6}{x}\) is a hyperbola with asymptotes at \(x = 0\) (vertical) and \(y = 0\) (horizontal). The graph of \(f(x) = \frac{6}{x-2}+1\) is a similar hyperbola but is translated 2 steps to the right and 1 step up, with asymptotes at \(x = 2\) (vertical) and \(y = 1\) (horizontal).

Step-by-step solution

Graph \(f(x)=\frac{6}{x}\)

Begin by graphing the function \(f(x)=\frac{6}{x}\). Instantly, it's clear that this is a hyperbola and it has vertical asymptote at \(x = 0\) because, as \(x\) approaches zero, the value of \(f(x)\) becomes very large. In the positive \(x\) region, the hyperbola will approach the \(x\)-axis but never touch or cross it; the same happens in the negative \(x\) region. Hence, there is a horizontal asymptote on the \(x\)-axis.

Graph \(f(x)=\frac{6}{x-2}+1\)

Next, graph the function \(f(x)=\frac{6}{x-2}+1\). This function resembles the first, but it has been moved 2 units to the right (because of the '-2' in the denominator) and one unit upward (because of the '+1' at the end). The vertical asymptote now will be at \(x = 2\) and the horizontal asymptote will be at \(y = 1\). Again, in the region \(x > 2\), the function will approach but never touch the line \(y = 1\), and also will happen in the region \(x < 2\).

Comparing graphs

Now compare the two graphs. Both functions are rational and they both yield a hyperbola. Nevertheless, the second function's graph is a translated version of the first: it moves 2 steps to the right, and 1 step up. Furthermore, the asymptotes' locations have been adjusted accordingly.