How do you find the asymptotes for #y=(9x)/(9x+3)#?

Answer 1

Cora Alexander

Horizontal asymptote y=1
Vertical asymptotex=-1/3

Write #y= 1 + (-3)/(9x+3)#

This shows one horizontal asymptote y=1 and one horizontal asymptote 9x+3=0, that is x= -1/3

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Answer 2

Cora Alexander

To find the asymptotes for $y = \frac{9x}{9x+3}$ , set the denominator equal to zero and solve for $x$ . The vertical asymptote occurs where the denominator equals zero.

$9x + 3 = 0$

$9x = -3$

$x = -\frac{1}{3}$

So, the vertical asymptote is $x = -\frac{1}{3}$ .

There are no horizontal or slant asymptotes in this case because the degree of the numerator is equal to the degree of the denominator. Therefore, the horizontal asymptote can be found by dividing the leading coefficient of the numerator by the leading coefficient of the denominator. In this case, it's $\frac{9}{9}$ which simplifies to $1$ . Therefore, the horizontal asymptote is $y = 1$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.