12
POINT OF DISCONTINUITY Notes 11.3 A point of discontinuity is the __________________ of a point where the graph of a function f(x) is ________ ______________________ 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. x-coordinate not continuous

POINT OF DISCONTINUITY Notes 11.3 A point of discontinuity is the __________________ of a point where the graph of a function f(x) is ________ ______________________

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POINT OF DISCONTINUITYNotes 11.3

A point of discontinuity is the __________________ of a point where the graph of a function f(x) is ________ ______________________

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.

x-coordinate

not continuous

REMOVABLE AND NON-REMOVABLEDISCONTINUITY

Removable Non-removable ________ 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

Holecontinuous

redefine

ORIGINALundefined

____________ in the graph There is no way to if

___________ the function at that point to make it continuous

Asymptote

redefine

By the way, continuous graphs have no jumps, breaks, holes or asymptotes!!

LETS EXPLORE HOW TO

FIND POINTS OF DISCONTINUITY

FIND POINT OF DISCONTINUITY : ASYMPTOTES

If Exponent is:

y=___ BOB0

BOTN

EATS DC

Y=0

No H.A.

Divide Coefficients of leading term

Simplify Function and set denominator equal to zero and solve for x

Bigger on bottom…0

Bigger on top…none

Exponents are the same

x=___Vertical Asymptote:

Horizontal Asymptote: Asymptote is

H.

Asymptote:

V. Asymptote:

24

1)(

x

xxf

y = 3

X = -4

X Y

-6

-5

-4

-3

-2

4.5

1.5

6

0

und

EATS DC

Set denominator equal to 0

Ex.

Divide coefficients and get 1 for the fraction but then add the constant of 2

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

FIND POINT OF DISCONTINUITY : HOLE(S)

1. Look at the given equation. 2.Factor the numerator and denominator3. If there is a common factor, set that factor equal to zero and solve for x.4.This x-value is where there will be a hole in the graph.

Graph)3(

)32( 2

x

xxy

)3(

)1)(3(

x

xx

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

1xEX:

This is removable discontinuity bc you can redefine the fcn at f(-3)=-4 and make the graph continuous

H. Asymptote:

Hole:

V. Asymptote:

4

122

2

x

xxy

y = 1

x = 2 and x = – 2

X Y

-4

-3

-1

0

1

3

4

4

3

2/3

0

10/3

EATS DC

Set denominator factors equal to 0

Ex

Divide leading coefficients and get 1 for the fraction

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

Factor to see if anything factors out.

)2)(2(

)3)(4(

xx

xx

Nothing factors out so there is no hole in the graph.

0

-1.2

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%?

solutionagetto

solutionofmL

%9.

%5.0275

x

xy

100

)005.0()02.0)(100(

xx 005.2)100(009.0

xx 005.2009.9.

1.1004. x

275x

You can also check this on the graphing calculator. Type the right side of equation into y1 and input 0.009 into y2. Use the intersection function (2nd calc 5) to solve for x.

H.

Asymptote:

y=x-5 R12

BOTN Divide using synthetic division

Oblique Asymptote(not tested)Opt

Bigger by one degree so there is an oblique asymptote

3

32)(

2

x

xxxf

1 – 2 – 3

–3

1

–3

–5

15

12y = x – 5 is the oblique asymptoteGraph on the TI84 to see what it looks like.

11.3 Vocabulary Support Attributes of Rational Functions Concept List

Choose the concept from the list above that best represents the item in each box.

1. the line that a graph approaches as y increases in absolute value

2. In the denominator, these reveal the points of discontinuity.

3. This type of discontinuity appears as a hole in the graph.

4. This type of graph has no jumps, breaks, or holes.

5. a function that you can write in the form

Vocab clarification : Match the following then discuss

HOMEWORK: WS 11.3