# How do you graph y=1/[x(x-2)]?

Jul 24, 2015

You find the intercepts and the asymptotes, and then you sketch the graph.

#### Explanation:

Step 1. Find the $y$-intercepts.

$y = f \left(x\right) = \frac{1}{x \left(x - 2\right)}$

$f \left(0\right) = \frac{1}{0 \left(0 - 2\right)} = \frac{1}{0}$

There is no $y$-intercept.

Step 2. Find the $x$-intercepts.

$0 = \frac{1}{x \left(x - 2\right)}$

$0 = 0$

There is no $x$-intercept.

Step 3. Find the vertical asymptotes.

Set the denominator equal to zero and solve for $x$.

$x \left(x - 2\right) = 0$

$x = 0$ or $x - 2 = 0$

$x = 0$ or $x = 2$

There are vertical asymptotes at $x = 0$ and $x = 2$.

Step 4. Find the horizontal asymptote.

The degree of the denominator is greater than the degree of the numerator, so

The horizontal asymptote is at $y = 0$ (the $x$-axis).

Step 5. Draw your axes and the asymptotes.

The vertical asymptotes divide the graph into three regions of $x$s.

(a) The left hand region has the $x$- and $y$-axes as asymptotes.

f(-1) = 1/(-1)(-1-2) = 1/(-1)(-3)) = 1/(3) ≈ 0.33.

The point at ($- 1 , 0.33$) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

(b) The right hand region has $x = 2$ and the $x$-axis as asymptotes.

f(3) = 1/(3(3-2)) = 1/(3×1) = 1/3 ≈ 0.33.

The point at ($3 , 0.33$) is in the second quadrant, so we have a "hyperbola" above the horizontal asymptote.

So we have a mirror-image hyperbola in the first quadrant.

(c) In the middle region, we have

$f \left(1\right) = \frac{1}{1 \left(1 - 2\right)} = \frac{1}{1 \left(- 1\right)} = \frac{1}{- 1} = - 1$ and

f(1.5) = 1/(1.5(1.5-2)) = 1/(1.5(-0.5)) = 1/(-0.75) ≈ -1.33

f(0.5) = 1/(0.5(0.5-2)) = 1/(0.5(-1.5)) = 1/(-0.75) ≈-1.33

The points at ($0.5 , - 1.33$), ($1 , - 1$), and ($1.5 , - 1.33$) are all below the $y$-intercept, so we have an "inverted parabola" between the vertical asymptotes.

And we have our graph.