Question

6. Let ()=45−63+2382−13+72, where N is a positive integer.A. Find all value(s) of N such that r(x) has a horizontal asymptote of y=0. B. Find all value(s) of N such that r(x) has a non-zero horizontal asymptote.C. Find the non-zero HA from part B. D. Find all value(s) of N such that r(x) has a slant asymptote. (You do not need to find these SAs).E. Find all values of N such that r(x) has no HA and no SA. F. Explain how you found your answers for A-E.

Answer 1

From the given function :

`r(x)=(4x^5-6x^3+23)/(8x^2-13x^N+72)`

The horizontal asymptotes depends on the coeficients of the highest degree between the numerator and the denominator.

Note that in any function :

`f(x)=(ax^n+\cdots)/(bx^m+\cdots)`

Where :

ax^n is the term with the highest degree in the numerator and

bx^m in the denominator.

If n < m :

The horizontal asymptote is y = 0

If n = m :

The horizontal asymptote is y = a/b

If n > m :

There is NO horizontal asymptote BUT there is a Slant Asymptote if n + 1 = m or n is exactly greater than m by 1.

Going back to the given function :

The term with the highest degree is 4x^5, so n = 5

For A. We need to find the values of N that will make the function have a horizontal asymptote of y = 0

From the note :

the values of n and m must be n < m

Since the value of n in the numerator is 5, the value of m in the denominator must be greater than 5

Assuming that the term -13x^N in the denominator has the highest degree.

N must be greater than 5

So N > 5

Now for B :

Values of N such that r(x) has a non-zero horizontal asymptote.

From the notes above, n must be equal to m (n = m).

Still, since the highest degree in the numerator is 5. The highest degree in the denominator must be equal to 5 also.

So N = 5

Then for C :

The horizontal asymptote from part B is y = a/b

a will be equal to 4

and

b will be equal to -13

So the asymptote in part B will be :

`\begin{gathered} y=(4)/(-13) \\ y=-(4)/(13) \end{gathered}`

For D :

So to have a slant asymptote, n must be greater than m by 1.

Still, the highest degree in the numerator is 5.

So the degree of the denominator must be 4 to satisfy the condition (n + 1 = m)

Taking again the term -13x^N

N must be equal to 4

N = 4

For E :

Values of N such that r(x) has no HA and SA.

The rule is n > m

Still, from the numerator, n = 5

and m must be less than 5 except 4.

(Because if the value of m is 4, it will have a Slant asymptote like in Letter D)

So we can say that m must be less than 4

Therefore, from the denominator :

N < 4 or N = 0, 1, 2 and 3