anonymous
  • anonymous
Suppose a function is defined as f(x) = [(x + a)(x + b)] divided by [(x + c)(x + d)]. What are its zeros, asymptotes, and holes? Now write a rational function with numbers instead of a, b, c, and d. Give the function's zeros, asymptotes, and holes. (Note that more than one rational function can have the same zeros, asymptotes, and holes. Therefore, many answers to this question are possible.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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terenzreignz
  • terenzreignz
Okay... zeros of a function (in particular, a rational one, such as this) would be the values of x that would make the NUMERATOR equal to zero. the numerator here is (x+a)(x+b) If we set (x+a)(x+b) = 0 What are the values of x?
anonymous
  • anonymous
1?
terenzreignz
  • terenzreignz
No....

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anonymous
  • anonymous
-x?
terenzreignz
  • terenzreignz
Okay... simply put (x+a)(x+b) = 0 only means either x+a = 0 or x+b = 0 Solve both these equations...
anonymous
  • anonymous
oh okay x+a-a=x x+b-b=x Like that?
terenzreignz
  • terenzreignz
=0 at the right sides, and not = x
anonymous
  • anonymous
Oh okay
terenzreignz
  • terenzreignz
Okay, like this... \[\large x+a = 0\] subtract a from both sides \[\large x+a\color{red}{-a}= 0 \color{red}{-a}\] simplify \[\large x = -a\] tadaa\ :D
anonymous
  • anonymous
Thanks! :D
terenzreignz
  • terenzreignz
That's not all of the answers yet...
terenzreignz
  • terenzreignz
Just one of them.
anonymous
  • anonymous
Oh okay, I'm listening. Go on
terenzreignz
  • terenzreignz
I really have to go now... I'm sure there are others here that can help you... The other equation is x+b=0 Would you please solve this?
terenzreignz
  • terenzreignz
(Not forgetting of course, that as of now, we are only solving for the zeros of the function)
anonymous
  • anonymous
x+b-b=0
terenzreignz
  • terenzreignz
you forgot to do the same to the right-side
anonymous
  • anonymous
0+b-b=0?
terenzreignz
  • terenzreignz
uhh... no... check the way I solved the x+a = 0 again
anonymous
  • anonymous
OH! x+b-b=0-b
terenzreignz
  • terenzreignz
Yes... continue...
anonymous
  • anonymous
x=-b
terenzreignz
  • terenzreignz
That's good. So, x = -b and x = -a are the zeros of the function. To find the asymptotes, this time, set the denominator (x+c)(x+d) to be equal to zero. Solve (x+c)(x+d) = 0
anonymous
  • anonymous
x+c-c=0-c x=-c x+d-d=0-d x=-d
terenzreignz
  • terenzreignz
That's good :) So, your asymptotes are x = -c and x = -d Next, the holes... They are the values which would make BOTH the numerator and denominator equal to zero. The zeros of the numerator are -a and -b The zeros of the denominator are -c and -d Do they have any in common?
anonymous
  • anonymous
they are negative?
terenzreignz
  • terenzreignz
No, I mean, does the numerator have any zeros that are also zeros of the denominator?
anonymous
  • anonymous
is it x?
terenzreignz
  • terenzreignz
No... what are the zeros of the numerator again?
anonymous
  • anonymous
-a and -b
terenzreignz
  • terenzreignz
And are any of these two also zeros of the denominator?
anonymous
  • anonymous
no
terenzreignz
  • terenzreignz
Then the function has no holes :P
anonymous
  • anonymous
ohh okay

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