Rational Functions

Question

The graph of the rational function (x-1)(x-1)/(x-1)(x+1) has a hole in the graph at (1, 0). My question: is 1 still considered the x-intercept? Our textbook says you should simplify the function and then look for zeros of the numerator. This makes it sound like it would be, but not 100% sure....

Answer 1

Yes please see below

Since the is actually fraction, you simplify it as such.

You can cancel out parts that are common to the numerator and denominator as follows

(x-1)(x-1)/(x-1)(x+1)

You can cancel out an (x-1) from the numerator and denominator

(x-1)(x-1)/(x-1)(x+1)

This leaves

(x-1)/(x+1) this is what you should be graphing

You have (x + 1) in the denominator remember the denominator cannot be zero so x cannot equal negative 1. Negative one is a boundary for this this function

If you set your function as f(x) or y = (x-1)/(x+1)

Set y = 0 then solve for x you will find your x intercept

0 = (x-1)/(x+1)

Multiply both sides by (x+1)

0 = (x-1)

Solve for x

x = 1

So when y is 0 x is 1.

You can graph your simplified function on Desmos.com

I hope you find this useful if you have questions please send me a message.

Answer 2

The function has two zeros in the denominator: x = 1 and x = -1. The zero (in the denominator) at x = 1 is removable, because the function can be reduced to (x-1)/(x+1).

Considering this function, there is an x intercept at x = 1 (1,0), because at x = 1, f(x) = (1-1)/(1+1) = 0/2 = 0