devalynn
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someone please help me ASAP. I’ll mark brainliest. thank you <3

For each function identify the degree of the numerator and the denominator.

1. f(x)=(x+5)(x-2)(x-7)/(x+9)(x-6)

2. h(x)=(6x+7)(x-4)/(3x+9) (x) (x^2)

3. g(x)=(5x+8)(x-4)/(3x+9) (x)

Which function in problems 1-3 has a horizontal asymptote at y=0?

Sagot :

sqdancefan

9514 1404 393

Answer:

  1. numerator: 3; denominator: 2; slant aymptote
  2. numerator: 2, denominator 4; y = 0 asymptote
  3. numerator 2, denominator: 2; y = 5/3 asymptote

Step-by-step explanation:

We have to assume you intend everything to the right of the slash (/) to be denominator. Ordinarily, only the first factor would be considered denominator according to the Order of Operations.

ab/cd = (ab/c)d . . . according to the order of operations.

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The degree of the numerator or denominator is essentially the number of factors involving x.

1. The numerator has 3 factors, hence degree 3. The denominator has 2 factors, hence degree 2. The degree of the numerator is 1 more than the degree of the denominator, so the function will have a slant asymptote.

  degrees: num/den = 3/2

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2. The numerator has 2 factors, hence degree 2. The denominator has one binomial factor and 3 factors of x, hence degree 4. The degree of the denominator is higher than the degree of the numerator, so the function will have y = 0 as its horizontal asymptote.

  degrees: num/den = 2/4 --- y = 0 horizontal asymptote

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3. The numerator has 2 factors, as does the denominator. Hence the degree of each is 2. The horizontal asymptote will be a constant equal to the ratio of the leading coefficients: y = 5/3.

  degrees: num/den = 2/2

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