# 2.8 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?

## Finding Zeroes of Rational Functions

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.

Take the following rational function:

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

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). Watch the video below and focus on the portion of this video discussing holes and $$x$$ -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 $$x$$ 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 $$x=1,2,3$$ and holes at $$x=0,4$$.

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:

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

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

### Example 3

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

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

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

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

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 4

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

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

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

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

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:

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

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

### Example 5

Identify the zeroes and holes of the following rational function.

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

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.

Review

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

1. $$f(x)=\frac{x^{3}+x^{2}-10 x+8}{x-2}$$

2. $$g(x)=\frac{6 x^{3}-17 x^{2}-5 x+6}{x-3}$$

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 $$x=3$$.

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