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Point of Discontinuity

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1 Point of Discontinuity
Notes 11.3 Point of Discontinuity A point of discontinuity is the __________________ of a point where the graph of a function f(x) is ________ ______________________ x-coordinate not continuous Points of discontinuity have already been introduced in this unit as asymptotes in the reciprocal function graph . In this section, we will look at many rational functions. We will identify additional points of discontinuity called “holes” in graphs.

2 Removable and non-removable discontinuity
________ in the graph Could be ________________ if _____________ the function at that point The hole occurs in the equation where there is a common factor in the numerator and denominator, thus the ______________ function is ______________ at a specific x-value Hole ____________ in the graph There is no way to if ___________ the function at that point to make it continuous Asymptote continuous redefine redefine By the way, continuous graphs have no jumps, breaks, holes or asymptotes!! ORIGINAL undefined

3 Lets explore how to find points of discontinuity

4 Find Point of Discontinuity : Asymptotes
Horizontal Asymptote: Asymptote is If Exponent is: BOB0 Y=0 y=___ Bigger on bottom…0 BOTN No H.A. Bigger on top…none EATS DC Divide Coefficients of leading term Exponents are the same Vertical Asymptote: Simplify Function and set denominator equal to zero and solve for x x=___

5 Ex. y = 3 X = -4 Pick 2 x-values to left and right of V.A.
H. Asymptote: V. Asymptote: EATS DC y = 3 Divide coefficients and get 1 for the fraction but then add the constant of 2 X Y -6 -5 -4 -3 -2 4.5 6 Set denominator equal to 0 und X = -4 1.5 Pick 2 x-values to left and right of V.A. This is non-removable discontinuity bc you can’t redefine the fcn and make the graph continuous

6 Find Point of Discontinuity : Hole(s)
1. Look at the given equation. Factor the numerator and denominator If there is a common factor, set that factor equal to zero and solve for x. This x-value is where there will be a hole in the graph.

7 Graph BOTN so no H.A. x+3 either causes a V.A. or a hole…
EX: Graph BOTN so no H.A. x+3 either causes a V.A. or a hole… Because x+3 factors out, there is a hole in the graph at x = – 3 This is removable discontinuity bc you can redefine the fcn at f(-3)=-4 and make the graph continuous

8 Ex y = 1 x = 2 and x = – 2 Pick x-values to left and right of V.A..
-4 -3 -1 1 3 4 Ex -1.2 H. Asymptote: Hole: V. Asymptote: 4 EATS DC 3 y = 1 10/3 Divide leading coefficients and get 1 for the fraction 2/3 Factor to see if anything factors out. Nothing factors out so there is no hole in the graph. Set denominator factors equal to 0 x = 2 and x = – 2 Pick x-values to left and right of V.A.. This is non-removable discontinuity bc you can’t redefine the fcn and make the graph continuous

9 The function below gives the concentration of the saline solution after adding x ml of the 0.5% solution to 100 ml of the 2% solution. How many mL of the 0.5% solution must you add for the combined solution to have a concentration of 0.9%? You can also check this on the graphing calculator. Type the right side of equation into y1 and input into y2. Use the intersection function (2nd calc 5) to solve for x.

10 Opt –3 1 – 2 – 3 y=x-5 R12 –3 15 y = x – 5 is the oblique asymptote 1
(not tested) Opt H. Asymptote: BOTN Divide using synthetic division Bigger by one degree so there is an oblique asymptote –3 – – 3 y=x-5 R12 –3 15 y = x – 5 is the oblique asymptote Graph on the TI84 to see what it looks like. 1 –5 12

11 Vocab clarification : Match the following then discuss

12 Homework: WS 11.3


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