Question

ap precalculus worksheet 1.8 name: period: find the zeros and undefined values for the following rational functions. factor where needed: 1. ( y = \frac{(x+3)(x-2)}{(x-1)(x+5)} ) zeros: ( x = ) ____ undef.: ( x = ) __ 2. ( y = \frac{x^2}{(x-1)(x+2)} ) zeros: ( x = ) __ undef.: ( x = ) __ 3. ( y = \frac{(x-3)(x-6)}{(x-6)(x-2)} ) zeros: ( x = ) __ undef.: ( x = ) __ 4. ( h(x) = \frac{x^2 - x - 20}{x^2 + x - 20} ) zeros: ( x = ) __ undef.: ( x = ) __ 5. ( f(x) = \frac{x^2 - 9}{x^2 - 2x - 15} ) zeros: ( x = ) __ undef.: ( x = ) __ 6. ( g(x) = \frac{x^3 - 4x^2 - 32x}{2x^2 + 7x - 4} ) zeros: ( x = ) __ undef.: ( x = ) ____

Explanation:

Problem 1

Step1: Find zeros (set numerator to 0)

\((x + 3)(x - 2)=0\) gives \(x=-3\) or \(x = 2\).

Step2: Find undefined (set denominator to 0)

\((x - 1)(x + 5)=0\) gives \(x = 1\) or \(x=-5\).

Step1: Find zeros (set numerator to 0)

\(x^{2}=0\) gives \(x = 0\).

Step2: Find undefined (set denominator to 0)

\((x - 1)(x + 2)=0\) gives \(x = 1\) or \(x=-2\).

Step1: Find zeros (set numerator to 0, cancel common factors)

Cancel \((x - 6)\) (note \(x
eq6\) from denominator). Then \((x - 3)=0\) gives \(x = 3\).

Step2: Find undefined (set denominator to 0)

\((x - 6)(x - 2)=0\) gives \(x = 6\) or \(x = 2\).

Answer:

Zeros: \(x=-3, 2\); Undef.: \(x = 1, -5\)

Problem 2