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# Zeroes of Rational Functions

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The zeroes of a function are the collection of  $x$ values where the height of the function is zero.  How do you find these values for a rational function and what happens if the zero turns out to be a hole?

#### Watch This

Focus on the portion of this video discussing holes and $x$ -intercepts.

http://www.youtube.com/watch?v=UnVZs2EaEjI James Sousa: Find the Intercepts, Asymptotes, and Hole of a Rational Function

#### Guidance

Zeroes are also known as $x$ -intercepts, solutions or roots of functions.  They are the $x$  values where the height of the function is zero.  For rational functions, you need to set the numerator of the function equal to zero and solve for the possible  $x$ values.  If a hole occurs on the $x$  value, then it is not considered a zero because the function is not truly defined at that point.

Example A

Identify the zeroes and holes of the following rational function.

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

Solution:  Notice how one of the $x+3$  factors seems to cancel and indicate a removable discontinuity.  Even though there are two  $x+3$ factors, the only zero occurs at $x=1$  and the hole occurs at (-3, 0).

Example B

Identify the zeroes, holes and  $y$ intercepts of the following rational function without graphing.

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

Solution:  The holes occur at $x=-1, 1$ .  To get the exact points, these values must be substituted into the function with the factors canceled.

$f(x) &= x(x-2)(x+1)(x+2)\\f(-1) &= 0, f(1)=-6$

The holes are (-1, 0); (1, 6).  The zeroes occur at $x=0, 2, -2$ .  The zero that is supposed to occur at $x=-1$  has already been demonstrated to be a hole instead.

Example C

Create a function with holes at $x=1,2,3$  and zeroes at $x=0, 4$

Solution:  There are an infinite number of possible functions that fit this description because the function can be multiplied by any constant.  One possible function could be:

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

Concept Problem Revisited

To find the zeroes of a rational function, set the numerator equal to zero and solve for the  $x$ values.  When a hole and a zero occur at the same point, the hole wins and there is no zero at that point.

#### Vocabulary

A zero is where a function crosses the $x$ -axis.  It is also known as a root, solution or $x$ -intercept.

A rational function is a function with at least one rational expression.

A rational expression is a ratio of two polynomial expressions.

#### Guided Practice

1.  Create a function with holes instead of zeroes.

2.  Identify the  $y$ intercepts, holes, and zeroes of the following rational function.

$f(x)=\frac{6x^3-7x^2-x+2}{x-1}$

3. Identify the zeroes and holes of the following rational function.

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

1.  There are an infinite number of functions that meet the requirements.  An illustrative example would be:

$f(x)=(x-1)(x+1) \cdot \frac{(x-1)(x+1)}{(x-1)(x+1)}$

The two  $x$ values that are holes are identical to the two  $x$ values that would be zeroes.  Therefore, this function has no zeroes because holes exist in their place.

2.  After noticing that a possible hole occurs at $x=1$  and using polynomial long division on the numerator you should get:

$f(x)=(6x^2-x-2) \cdot \frac{x-1}{x-1}$

A hole occurs at $x=1$  which turns out to be the point (1, 3) because $6 \cdot 1^2-1-2=3$

The $y$ -intercept always occurs where $x=0$  which turns out to be the point (0, -2) because $f(0)=-2$ .

To find the $x$ -intercepts you need to factor the remaining part of the function:

$(2x+1)(3x-2)$

Thus the zeroes ( $x$ -intercepts) are $x=-\frac{1}{2}, \frac{2}{3}$ .

3. The hole occurs at $x=-1$  which turns out to be a double zero.  The hole still wins so the point (-1, 0) is a hole.  There are no zeroes.  The constant 2 in front of the numerator and the denominator serves to illustrate the fact that constant scalars do not impact the  $x$ values of either the zeroes or holes of a function.

#### Practice

Identify the intercepts and holes of each of the following rational functions.

1. $f(x)=\frac{x^3+x^2-10x+8}{x-2}$
2. $g(x)=\frac{6x^3-17x^2-5x+6}{x-3}$
3. $h(x)=\frac{(x+2)(1-x)}{x-1}$
4. $j(x)=\frac{(x-4)(x+2)(x+2)}{x+2}$
5. $k(x)=\frac{x(x-3)(x-4)(x+4)(x+4)(x+2)}{(x-3)(x+4)}$
6. $f(x)=\frac{x(x+1)(x+1)(x-1)}{(x-1)(x+1)}$
7. $g(x)=\frac{x^3-x^2-x+1}{x^2-1}$
8. $h(x)=\frac{4-x^2}{x-2}$
9. Create a function with holes at $x=3, 5, 9$  and zeroes at $x=1, 2$ .
10. Create a function with holes at $x=-1, 4$  and zeroes at $x=1$
11. Create a function with holes at $x=0, 5$  and zeroes at $x=2, 3$ .
12. Create a function with holes at $x=-3, 5$  and zeroes at $x=4$
13. Create a function with holes at $x=-2, 6$  and zeroes at $x=0, 3$ .
14. Create a function with holes at $x= 1, 5$  and zeroes at $x=0,6$ .
15. Create a function with holes at $x=2, 7$  and zeroes at $x=3$ .