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# Zeroes of Rational Functions

## Values where the numerator equals zero but the denominator doesn't.

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Zeroes of Rational Functions

The zeroes of a function are the collection of x\begin{align*}x\end{align*} 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

### Finding Zeroes of Rational Functions

Zeroes are also known as x\begin{align*}x\end{align*}-intercepts, solutions or roots of functions.  They are the x\begin{align*}x\end{align*} 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\begin{align*}x\end{align*} values.  If a hole occurs on the x\begin{align*}x\end{align*} value, then it is not considered a zero because the function is not truly defined at that point.

Take the following rational function:

f(x)=(x1)(x+3)(x+3)x+3\begin{align*}f(x)=\frac{(x-1)(x+3)(x+3)}{x+3}\end{align*}

Notice how one of the x+3\begin{align*}x+3\end{align*} factors seems to cancel and indicate a removable discontinuity.  Even though there are two x+3\begin{align*}x+3\end{align*} factors, the only zero occurs at x=1\begin{align*}x=1\end{align*} and the hole occurs at (-3, 0).

Watch the video below and focus on the portion of this video discussing holes and x\begin{align*}x\end{align*}-intercepts.

### Examples

#### Example 1

Earlier, you were asked how to find the zeroes of a rational function and what happens if the zero is a hole. To find the zeroes of a rational function, set the numerator equal to zero and solve for the x\begin{align*}x\end{align*} values. When a hole and a zero occur at the same point, the hole wins and there is no zero at that point.

#### Example 2

Create a function with zeroes at x=1,2,3\begin{align*}x=1,2,3\end{align*} and holes at x=0,4\begin{align*}x=0, 4\end{align*}

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)=(x1)(x2)(x3)x(x4)x(x4)\begin{align*}f(x)=\frac{(x-1)(x-2)(x-3)x(x-4)}{x(x-4)}\end{align*}

Note that 0 and 4 are holes because they cancel out.

#### Example 3

Identify the zeroes, holes and y\begin{align*}y\end{align*} intercepts of the following rational function without graphing.

f(x)=x(x2)(x1)(x+1)(x+1)(x+2)(x1)(x+1)\begin{align*}f(x)=\frac{x(x-2)(x-1)(x+1)(x+1)(x+2)}{(x-1)(x+1)}\end{align*}

The holes occur at x=1,1\begin{align*}x=-1, 1\end{align*}.  To get the exact points, these values must be substituted into the function with the factors canceled.

f(x)f(1)=x(x2)(x+1)(x+2)=0,f(1)=6\begin{align*}f(x) &= x(x-2)(x+1)(x+2)\\ f(-1) &= 0, f(1)=-6\end{align*}

The holes are (-1, 0); (1, 6).  The zeroes occur at x=0,2,2\begin{align*}x=0, 2, -2\end{align*}.  The zero that is supposed to occur at x=1\begin{align*}x=-1\end{align*} has already been demonstrated to be a hole instead.

#### Example 4

Identify the y\begin{align*}y\end{align*} intercepts, holes, and zeroes of the following rational function.

f(x)=6x37x2x+2x1\begin{align*}f(x)=\frac{6x^3-7x^2-x+2}{x-1}\end{align*}

After noticing that a possible hole occurs at x=1\begin{align*}x=1\end{align*} and using polynomial long division on the numerator you should get:

f(x)=(6x2x2)x1x1\begin{align*}f(x)=(6x^2-x-2) \cdot \frac{x-1}{x-1}\end{align*}

A hole occurs at x=1\begin{align*}x=1\end{align*} which turns out to be the point (1, 3) because 61212=3\begin{align*}6 \cdot 1^2-1-2=3\end{align*}

The y\begin{align*}y\end{align*}-intercept always occurs where x=0\begin{align*}x=0\end{align*} which turns out to be the point (0, -2) because f(0)=2\begin{align*}f(0)=-2\end{align*}.

To find the x\begin{align*}x\end{align*}-intercepts you need to factor the remaining part of the function:

(2x+1)(3x2)\begin{align*}(2x+1)(3x-2)\end{align*}

Thus the zeroes (x\begin{align*}x\end{align*}-intercepts) are x=12,23\begin{align*}x=-\frac{1}{2}, \frac{2}{3}\end{align*}.

#### Example 5

Identify the zeroes and holes of the following rational function.

f(x)=2(x+1)(x+1)(x+1)2(x+1)\begin{align*}f(x)=\frac{2(x+1)(x+1)(x+1)}{2(x+1)}\end{align*}

The hole occurs at x=1\begin{align*}x=-1\end{align*} 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\begin{align*}x\end{align*} values of either the zeroes or holes of a function.

### Review

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

1. f(x)=x3+x210x+8x2\begin{align*}f(x)=\frac{x^3+x^2-10x+8}{x-2}\end{align*}
2. g(x)=6x317x25x+6x3\begin{align*}g(x)=\frac{6x^3-17x^2-5x+6}{x-3}\end{align*}
3. h(x)=(x+2)(1x)x1\begin{align*}h(x)=\frac{(x+2)(1-x)}{x-1}\end{align*}
4. j(x)=(x4)(x+2)(x+2)x+2\begin{align*}j(x)=\frac{(x-4)(x+2)(x+2)}{x+2}\end{align*}
5. k(x)=x(x3)(x4)(x+4)(x+4)(x+2)(x3)(x+4)\begin{align*}k(x)=\frac{x(x-3)(x-4)(x+4)(x+4)(x+2)}{(x-3)(x+4)}\end{align*}
6. f(x)=x(x+1)(x+1)(x1)(x1)(x+1)\begin{align*}f(x)=\frac{x(x+1)(x+1)(x-1)}{(x-1)(x+1)}\end{align*}
7. g(x)=x3x2x+1x21\begin{align*}g(x)=\frac{x^3-x^2-x+1}{x^2-1}\end{align*}
8. h(x)=4x2x2\begin{align*}h(x)=\frac{4-x^2}{x-2}\end{align*}
9. Create a function with holes at x=3,5,9\begin{align*}x=3, 5, 9\end{align*} and zeroes at x=1,2\begin{align*}x=1, 2\end{align*}.
10. Create a function with holes at x=1,4\begin{align*}x=-1, 4\end{align*} and zeroes at x=1\begin{align*}x=1\end{align*}
11. Create a function with holes at x=0,5\begin{align*}x=0, 5\end{align*} and zeroes at x=2,3\begin{align*}x=2, 3\end{align*}.
12. Create a function with holes at \begin{align*}x=-3, 5\end{align*} and zeroes at \begin{align*}x=4\end{align*}
13. Create a function with holes at \begin{align*}x=-2, 6\end{align*} and zeroes at \begin{align*}x=0, 3\end{align*}.
14. Create a function with holes at \begin{align*}x= 1, 5\end{align*} and zeroes at \begin{align*}x=0,6\end{align*}.
15. Create a function with holes at \begin{align*}x=2, 7\end{align*} and zeroes at \begin{align*}x=3\end{align*}.

To see the Review answers, open this PDF file and look for section 2.8.

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### Vocabulary Language: English

TermDefinition
Hole A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero.
Rational Expression A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Function A rational function is any function that can be written as the ratio of two polynomial functions.
Zero The zeros of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.