How do you identify all asymptotes or holes for #f(x)=(x^29)/(2x^2+1)#?
horizontal asymptote at
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is nonzero for these values then they are vertical asymptotes.
there are no real solutions for x hence there are no vertical asymptotes.
Horizontal asymptotes occur as
Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence there are no holes. graph{(x^29)/(2x^2+1) [18.01, 18.04, 9.02, 9]}
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To identify all asymptotes or holes for ( f(x) = \frac{x^2  9}{2x^2 + 1} ), follow these steps:

Check for Vertical Asymptotes: Set the denominator equal to zero and solve for ( x ). Any ( x )value that makes the denominator zero will result in a vertical asymptote. In this case, the denominator ( 2x^2 + 1 ) has no real roots, so there are no vertical asymptotes.

Check for Horizontal Asymptotes: Compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at ( y = 0 ). If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote. Here, since the degree of the numerator is equal to the degree of the denominator, we look at the ratio of the leading coefficients. The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is 2. Therefore, the horizontal asymptote is ( y = \frac{1}{2} ).

Check for Oblique (Slant) Asymptotes (if applicable): If the degree of the numerator is exactly one more than the degree of the denominator, divide the numerator by the denominator to find the oblique asymptote. In this case, since the degree of the numerator is not one more than the degree of the denominator, there are no oblique asymptotes.

Check for Holes: Factor both the numerator and the denominator completely and cancel out any common factors. If there are any common factors, they represent potential holes in the graph. Here, the numerator ( x^2  9 ) factors into ( (x + 3)(x  3) ), and the denominator ( 2x^2 + 1 ) does not factor further. There are no common factors to cancel out, so there are no holes in the graph.
Therefore, the function ( f(x) = \frac{x^2  9}{2x^2 + 1} ) has a horizontal asymptote at ( y = \frac{1}{2} ), but no vertical asymptotes, oblique asymptotes, or holes.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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