How do you find the Vertical, Horizontal, and Oblique Asymptote given #g(x)= (-2x+3) /( 3x+1)#?

Answer 1

horizontal #y = -2/3#

vertical #x = -1/3#

testing for horizontal or oblique

#(-2x+3)/(3x+1) ]_{x \to pm \infty} #
#= (-2+3/x)/(3+1/x) ]_{x \to pm \infty} = -2/3#

for vertical , we look at denominator = 0

so #3x+1 = 0 \implies x = -1/3 #
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Answer 2

To find the vertical asymptote, set the denominator equal to zero and solve for x. For the function g(x) = (-2x + 3) / (3x + 1), the vertical asymptote occurs at x = -1/3.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (-2x + 3) is less than the degree of the denominator (3x + 1), the horizontal asymptote is y = 0.

To find the oblique asymptote, perform long division or polynomial division between the numerator and denominator. After division, the quotient will represent the oblique asymptote. In this case, performing long division yields y = -2/3x + 3.

So, the vertical asymptote is x = -1/3, the horizontal asymptote is y = 0, and the oblique asymptote is y = -2/3x + 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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