Graph .

Find any asymptotes by checking for which x-values the denominator is equal to zero.

In this case, f(x) is undefined for x = 2. Boom, that means x = 2 is an asymptote.

Now for the intercepts. We find the y-intercept by evaluating f(0).

So the y-intercept is (0, -1).

We find the x-intercept by setting the numerator equal to zero. But in this case, the numerator is always just 2. So there ain't no x-intercept.

But once again the x-axis acts as a horizontal asymptote.The graph will approach it but never cross it.

Let's make a table to get a few more points.

Sketch the axes and draw in a dotted vertical line at x = 2. Then plot your points using the x-axis and x = 2 as asymptotes.

Graph .

All right, let's kick things off with an asymptote party. Hmm...it looks like f(x) is undefined for:

x² + 2x = 0

Factor solve to find the zeros.

x(x + 2) = 0

That means x = 0 and x = -2, so we've got two vertical asymptotes at x = 0 and x = -2.

Since x = 0 is an asymptote, we have no y-intercept.

Once again we can't set the numerator equal to zero, so there's no x-intercept, but the x-axis does act as a horizontal asymptote. Now we need a table to get a few points.

Sketch the axes and draw in a dotted vertical line at x = -2. Then plot the points from the table and sketch in a graph, using the x-axis, x = 0, and x = -2 as asymptotes.

Graph .

Our denominator is obviously gonna be zero when x = 0, so we've got a vertical asymptote at x = 0. That means the entire y-axis is an asymptote, so there's no y-intercept.

Since our numerator is just 4, there's also no x-intercept.

Looks like we're intercept-free. Bummer. But we do know the x-axis acts as a horizontal asymptote.The graph will approach it but never cross it.

Let's make a table to get a few more points.

Sketch the axes. Plot your points using the x-axis and y-axis as asymptotes.