Question

Which function has a removable discontinuity

Answer 1

Answer:

$\frac{x-2}{x^2-x-2}$

Step-by-step explanation:

A removable discontinuity is when there is a hole in your graph. This is usually because one X value has been canceled out. Most of the time, it takes factoring to figure out if there is a removable discontinuity when looking at an equation.

First, look at the numerator $x-2$ . This can't be factored any further. However, $x^2-x-2$ can be factored since it is a trinomial (has three terms) .

For the purposes of this example, you may want to think about it as

$1x^2 -1x-2$

To factor, multiply the the outside coefficients

1 x -2 = -2

Now take the middle coefficient (-1) and ask yourself what two numbers multiply to make -2, but still add to be -1.

-2 x 1 = -2

-2 + 1 = -1

So in factored form, the equation is

$\frac{x-2}{(x-2)(x+1)}$

Since you have x-2 on both top and bottom, that can be canceled out. x - 2 would be your removable discontinuity in this situation.