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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)}

Answers:

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 .

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