Solving Discontinuities in Rational Equations: Factoring & Identifying Asymptotes, Exercises of Mathematics

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Solving for

Discontinuities

Algebraically

16 – 17 November 2010

Always Factor!  The 1st^ step → always factor the numerator and the denominator!!!  Goal: Get matching factors in numerator and denominator 1 ( 5 )( 1 ) 1 4 5

        x x x y x x x y

Example:

x

x

x

x x

y

x

x x

y

2

Your Turn:  Complete problems 1 – 5 on the “Solving for the Discontinuities of Rational Equations” handout.

Removable Discontinuities, cont.  Step 1: Factor the numerator and the denominator  Step 2: Identify factors that occur in both the numerator and the denominator  Step 3: Set the common factors equal to zero  Step 4: Solve for x  Step 5: Write your answers in the form x =

Example: ( 2 ) ( 2 )( 2 ) 2 4 2        x x x y x x y : 2 2 0     Hole x x

Vertical Asymptote vs. Removable Discontinuity  Algebraically, they act similarly  Consider:

( 2 ) ( 2 )( 2 ) ( 2 ) 4 4         x x x y x x x y

Vertical Asymptote vs. Removable Discontinuity, cont. 3 3 2 ( 2 )

x x x y x x x y !!! 0 0 0 4 8 4 ( 2 2 ) ( 2 ) 4 ( 2 ) 4 2 3 3 2              y y y x

Vertical Asymptote vs. Removable Discontinuity, cont. ( 2 )( 2 )( 2 ) ( 2 )( 2 ) ( 2 ) ( 2 )( 2 ) ( 2 ) 4 4 3 3 2               x x x x x y x x x y x x x y : 2 2 0 2 1       VA x x x y

Vertical Asymptote vs. Removable Discontinuity, cont.  Depends on:  (^) How many times a factor occurs  Where the factor occurs  Removable Discontinuity → the multiplicity of the factor in the numerator ≥ the multiplicity of the factor in the denominator  Vertical Asymptote → the multiplicity of the factor in the numerator < the multiplicity of the factor in the denominator

Your Turn:  Complete problems 11 – 15 on the “Solving for the Discontinuities of Rational Equations” handout.

Homework  In Precalculus textbook, pg. 290: 7 – 12  Hint! You will need to use the quadratic formula for #8.

Example 1  If n = m → HA:  If n < m → HA: y = 0  If n > m → HA doesn’t exist b a y 

x

x

y

HA : y  0

Example 2  If n = m → HA:  If n < m → HA: y = 0  If n > m → HA doesn’t exist b a y  2 7 13

    x x x y HA: none