How do you find the asymptotes for #f(x)=(2x)/(sqrt (9x^2-4))#?

Answer 1

Asymptotes: f(x) = y = #+-2/3# and #x=+-2/3#..

Let y = f(x). It is defined for #9x^2-4>0#. So, #|x|>2/3#.
#yto+-oo# as #xto+-2/3#. So, #x=+-2/3# are asymptotes..
Inversely, solve for x. #x=(2y)/sqrt(9y^2-4)#. Interestingly the form is the same. So, #|y|>2/3#.
#xto+-oo# as #yto+-2/3#. So, #y=+-2/3# are asymptotes..

The graph comprises four branches symmetrical about the origin in four regions:

Both x and #y > 2/3# in the first quadrant, #x<-2/3# and #y > 2/3# in the second quadrant, both x and #y < -2/3# in the third quadrant and #x > 2/3# and #y < -2/3# in the fourth quadrant.
As a whole, the graph is outside of #|x| <=2/3 and |y|<=2/3#.
In this formula, #f^(-1)-=f#, like f(x) = 1/x. Indeed, interesting.
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Answer 2

To find the asymptotes of the function ( f(x) = \frac{2x}{\sqrt{9x^2 - 4}} ), we need to consider the behavior of the function as ( x ) approaches certain values.

  1. Vertical Asymptotes: Vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. In this case, ( 9x^2 - 4 ) cannot equal zero, so we solve ( 9x^2 - 4 = 0 ) to find the vertical asymptotes.

    [ 9x^2 - 4 = 0 ] [ 9x^2 = 4 ] [ x^2 = \frac{4}{9} ] [ x = \pm \frac{2}{3} ]

    Thus, the vertical asymptotes occur at ( x = \frac{2}{3} ) and ( x = -\frac{2}{3} ).

  2. Horizontal Asymptotes: Horizontal asymptotes occur when ( x ) approaches positive or negative infinity. To find them, we examine the behavior of the function as ( x ) approaches infinity and negative infinity.

    As ( x ) approaches infinity, the terms with the highest power dominate the function's behavior. In this case, the highest power term is ( x ) in both the numerator and the denominator. Therefore, we can use the ratio of leading coefficients to find the horizontal asymptote.

    [ \lim_{x \to \infty} \frac{2x}{\sqrt{9x^2 - 4}} ]

    Since ( x ) is in the numerator and the denominator, the leading coefficients are 2 and 9. Therefore, the horizontal asymptote is ( y = \frac{2}{3} ).

    Similarly, as ( x ) approaches negative infinity, the horizontal asymptote is also ( y = \frac{2}{3} ).

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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.

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.

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.

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.

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