Question

How do you find the horizontal asymptote of f(x)= sqrt(36x^2+7)/9x+4?

Answer 1

Make x very large and see what the function becomes. For example, 9x +4 just becomes 9x.

Answer 2

My solution manual says for x>0, \[x ^{-1}\]= \[\left| x ^{-1} \right|\] = \[\sqrt{x^{-1}}\], and I have no clue why and how it equals that..

Answer 3

oops i meant \[\sqrt{x^{-2}}\]

Answer 4

Yes, in general, sqrtx^2n) = x^n, as taking square root cuts exponent in half.

So when x gets very large 36x^2 + 7 becomes nearly 36x^2.

Now we have sqrt(36 x^2) / 9 x for large x,

which becomes 6 x / 9 x after taking the square root

or f(x) goes toward 6/9 = 2/3, which is the answer.

Answer 5

But why/how can you say that 36x^2 + 7 becomes nearly 36x^2? Do you assume that or do you get that from some specific part of the problem?

Answer 6

f(x)= sqrt(36x^2+7)/9x+4

I presume you mean f(x)= sqrt(36x^2+7)/(9x+4) -- That is NOT what you wrote.

Anyway, what is the horizontal asymptote of this? \(g(x) = \dfrac{6x+5}{9x+4}\)

Answer 7

Tk, I don't understand what you mean, that IS what I wrote at the top. Also, the answer in the solution manual is 2/3 and -2/3.

Answer 8

1) You should be learning, not arguing. That is NOT what you wrote at the top. Please remember your Order of Operations and use parentheses to clarify intent.

1/9x+4 = \(\dfrac{1}{9x} + 4\) or, there is an argument that it is \(\dfrac{1}{9}x + 4\)

1/(9x+4) = \(\dfrac{1}{9x+4}\) The parentheses are NOT optional if this is what you mean.

2) You didn't answer my question.

Answer 9

@cassieWOAH Soluton manual is right, I forgot to check for large negative x values, which gives an asymptote at -2/3 (substitute large -x into equation).

Answer 10

If you answer my questions, we discover the meanings hidden in the problem statements.