Graphing - Part III

@Prepared for Gina. L

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Question 1: Find domain, x-intercept points, y-intercept points, symmetry, all the asymptotes 

and find the actual function base on the given graph (shown on Figure#1).

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                                      Figure#1

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1 - Domain 

Base on Figure#1, we can easily see that there are two vertical asymptotes at x = 1 and x = −1,

therefore,  Domain (, −1)(−1, 1)(1, )

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2 - X- intercepts points

Base on Figure#1, the x-int points are (2, 0) and (−2, 0) 

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3 - Y- intercepts points

Base on Figure#1, the y-int point is (0, 4)

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4 - Symmetric

There is symmetric about the y-axis, this is an even function, y(x) = y(−x).

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5 - Asymptotes

Two vertical asymptotes, x = 1 and x = −1.

One horizontal asymptotes, y = 1.

No slant asymptotes (oblique asymptotes)

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6 - Actual Function

6-1

Let y = f(x).

from the x-intercepts are (2, 0) and (−2, 0), which f(2) = 0, f(−2) = 0,

therefore, the "numerator must exist (x + 2) and (x − 2) terms" --- (requirement 1).  

we can currently assume that f(x) = (x + 2)(x − 2).

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p.s. f(x) = n(x)/d(x), n(x) is the numerator, and d(x) is denominator.

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example

if x-intercepts are (0, 0), (1, 0) and (2, 0), 

then the possible f(x) = (x)(x − 1)(x − 2) = x(x2 − 3x + 2) = x3 − 3x2 + 2x

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6-2

from the y-intercepts (0, 4)

"f(0) = 4 must be satisfy" --- (requirement 2)

however, base on our pervious assumption,   f(0) = (0+2)(0−2) = −4,  which is missing a minus sign (−).

now, we currently assume that f(x) = (x + 2)(x − 2).

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6-3

base on the VA (vertical asymptotes), there are two poles at x=1, and x = −1.

poles goes to the denominator of a rational function, (due to the zero on the denominator)

therefore, the "denominator must exist (x+1) and (x −1) terms" --- (requirement 3) .

again, we can now assume the possible function is...

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f(x) = (x + 2)(x − 2)

           (x + 1)(x − 1)

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example:

if VA is x = 100, then the possible f(x) = 1/(x100), which x cannot equal 100.

if VA is x = −100, then the possible f(x) = 1/(x+100), which x cannot equal −100.

if VA are x = 3 and x = −5, then the possible f(x) = 1/((x −3)(x+5)) = 1/(x2 + 2x − 15) , which x cannot equal 3 or −5.

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6-4

base on the HA (horizontal asymptotes), which is y = 1,

   lim         f(x)  =  constant (asymptotes)

  x->

Usually, the horizontal asymptotes exist only when the highest degree of numerator < the highest degree of denominator.

since we have a horizontal asymptotes at y = 1, so "  lim    f(x) =    lim     f(x) = 1 must be satisfy" --- (requirement 4) 

@                                                             x->                  x->−                                                  

therefore, 

1 ?= lim f(x)   = lim     (x + 2)(x − 2)

        x->          x->    (x + 1)(x − 1)

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                    = lim         −x^2 + 4   1/x^2

                       x->         x^2 −1     1/x^2

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                     = lim        1 + 4/x^2

                        x->         1 −1/x^2

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                    =    1 + 0

                             1 − 0

 

                    = −1

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Therefore, we missing a minus sign in the f(x), therefore, add a minus sign on our assumption from 6-3, 

f(x) = −  −(x + 2)(x − 2)  =  (x + 2)(x − 2) 

               (x + 1)(x − 1)        (x + 1)(x − 1)

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6-5

The last step, we shall re-check all four requirements.  They MUST all be satisfy. 

base on our assumption from 6-4, double check the following... 

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"numerator must exist (x + 2) and (x − 2) terms" --- (requirement 1)     ... Satisfy

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"f(0) = 4 must be satisfy" --- (requirement 2)                                      ... Satisfy

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"denominator must exist ( x+1) and ( x −1) terms" --- (requirement 3)    ... Satisfy

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"  lim    f(x) =    lim     f(x) = 1 must be satisfy" --- (requirement 4)        ... Satisfy

   x->                  x->−                                                    

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Therefore, our assumption on 6-4 is correct

f(x) = (x + 2)(x − 2)

          (x + 1)(x − 1)

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     =  x^2 − 4

         x^2 − 1

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6-6 Check 

We can check our answer with any online function graphing tool or MATLAB,

type (x^2 - 4)/(x^2 - 1) into the tool, we get the exact same result as question (shown on Figure#2).

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                                                                   Figure #2 

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p.s.

Online Graphing  

http://www.walterzorn.com/grapher/grapher_e.htm

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MATLAB

http://www.mathworks.com/

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Question 2: Find domain, x-intercept points, y-intercept points, symmetry, all the asymptotes 

and find the actual function base on the given graph (shown on Figure#3).

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                                           Figure#3

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1 - Domain 

Base on Figure#3, we can easily see is continuous for all the value of x from − to ,

therefore,  Domain (−, )

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2 - X- intercepts points

Base on Figure#3, the x-int points are (0,0), (2, 0) and (−2, 0) 

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3 - Y- intercepts points

Base on Figure#3, the y-int point is (0, 0)

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4 - Symmetric

There is symmetric about the y-axis, this is an even function, y(x) = y(−x).

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5 - Asymptotes

No vertical asymptotes.

No horizontal asymptotes.

No slant asymptotes (oblique asymptotes).

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6 - Actual Function

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6-1

Let y = f(x)

From the information on x-intercepts points, 

1st, we can see that the function bounce back at (0, 0), it is a "double roots", therefore, x2 must be a term of f(x).

2nd, we can see that the function pass through at (2, 0) and (−2, 0), they are "single root", therefore, (x + 2) and (x−2), must be a term of f(x).

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So, f(x) = kx2(x + 2)(x−2), "k" is a unknown number that we need to find out.

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6-2

The information of y-intercept (0, 0) is useless, because it is the same point as one of the x-intercepts point.

we can do nothing here, just move on.

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6-3

There are one max point (0, 0) and two min points (?,−2) and (?, −2), but we are missing the exact value of x, 

at these three points, f '(x) = 0 due to the tangent slope equal zero.

This is a very useful information, which can allowed us to find useful point on the function, and lead us to find the value of "k".

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f '(x) = d/dx kx2(x + 2)(x−2)

        = k d/dx x2(x2 −4)

        = k d/dx (x4 −4x2)

        = k (4x3 −8x)

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set f '(x) = 0

0 = k (4x3 −8x)

0 = 4x (x2 −2)

x = 0, 2, −2    therefore, the three points are --> (0, 0) (−2, −2) and (2, −2)

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(0, 0) is a useless point, since it the same point as x-int and y-int, 

however, we can use (−2, −2) and (2, −2) to find "k",

by sub the point (2, −2) into our function from 6-1, f(x) = kx2(x + 2)(x−2)

 

f(2) = k(2)2(2 + 2)(2−2)

          = k (2)(2 − 4)

          = −4k

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which f(2) = −2

so       −2 = −4k

             k = 1/2

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Our final answer will be f(x) = (1/2) x2(x + 2)(x−2)

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Answer: f(x) = (1/2) x2(x + 2)(x−2)

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6-4 Check

Type (1/2)*(x^2)*(x+2)*(x-2) into online graphing tool or MATLAB, 

we can see it is exact the same as our question (shown on Figure #4)

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                                                                   Figure #4 

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Written by Kevin Tang (Apr 2, 2010)

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