In the rational function $f(x)=\frac{p(x)}{q(x)} where q(x)\neq 0$ , when the factors of p(x) and q(x) are same then there is a hole in a rational function at that value of x.

Sometimes, same factor may appear in both the numerator and denominator. Let us consider that the factor $(x – a)^{n}$ is in the numerator and $(x – a)^{m}$ is in the denominator.

When n = m, then there will be a hole in the graph at x = a,the hole will not be on the x-axis. The y-value of the hole can be found by canceling the factors and substituting x = a in the reduced function.

For example : $f(x)=\frac{x^3(x-3)}{(x-3)} $

The factor (x-3) in p(x) and q(x) are same so there is a hole.

x -3 = 0

∴ x= 3

So the given rational function has a hole at x= 3.

$y = \frac{x^2+x-6}{-4x^2-16x-12} $

Factorize the numerator and denominator

$y= \frac{(x+3)(x-2}{-4(x+3)(x+1)} $

As the factor x+3 is present in both numerator and the denominator so there is a hole.

x + 3 = 0

∴ x= -3

So the hole of this rational function is at x = -3

2) Identify the holes in the given rational function if any. $f(x)=\frac{1}{3x^2+3x-18} $

$y = \frac{1}{3x^2+3x-18} $

Factorize the denominator since the numerator contains only 1

$y= \frac{1}{3(x+3)(x-2)} $

As there is no common factor in numerator and the denominator so there is no hole.

So the hole of this rational function has no hole.

3) Identify the hole from the given graph.

4) Identify the holes in the given rational function if any. $f(x)=\frac{x-4}{-4x-16}$

$y = \frac{x-4}{-4x-16} $

Factorize the denominator since we can not factorize the numerator

$y= \frac{x-4}{-4(x+4)} $

As there is no common factor in numerator and the denominator so there is no hole.

So the hole of this rational function has no hole.

1) $f(x)=\frac{x+2}{2x+3}$

2) $f(x)=\frac{x-2}{x^2-4}$

3) $f(x)=\frac{x^3 -9x}{3x^2-6x-9}$

precalculus

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