How do you find the asymptote(s) or hole(s) of #f(x) = (x+3)/(x^29)#?
VA:
HA:
We have the following expression
What I have in blue is a Difference of Squares of the form
Now, our new expression is
which simplifies to
Vertical asymptotes are found by setting the denominator equal to zero, because this is the value for which the function is undefined.
Horizontal asymptotes are found by comparing the degrees. Since there is a higher degree in the denominator, we have a horizontal asymptote at
Hope this helps!
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asymptote at
hole at
Factor the denominator, it is the difference of squares:
graph{(x+3)/(x^29) [6.67, 13.33, 4.44, 5.56]}
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To find the asymptotes and holes of (f(x) = \frac{x+3}{x^29}), follow these steps:

Factor the denominator and numerator if possible: [ f(x) = \frac{x+3}{(x3)(x+3)} ]

Identify and cancel common factors: The numerator and denominator share a common factor of (x+3). When you cancel this factor, you're left with: [ f(x) = \frac{1}{x3} ] The cancellation of the (x+3) term suggests there's a hole in the graph at (x = 3).

Find the hole's ycoordinate: Substitute (x = 3) into the simplified function (f(x) = \frac{1}{x3}) to find the ycoordinate of the hole. Since the factor was canceled and it's no longer part of the function, this step is more about recognizing that we had this term before simplification. The hole is at (x = 3), but we refer to the original function before cancellation for the yvalue: [ y = \frac{1}{3  3} = \frac{1}{6} ] So, the hole is at ((3, \frac{1}{6})).

Determine vertical asymptotes: After cancellation, vertical asymptotes occur where the denominator of the simplified function is zero (if any factors are left in the denominator after cancellation). For (f(x) = \frac{1}{x3}), set the denominator equal to zero and solve for (x): [ x3 = 0 \Rightarrow x = 3 ] So, there's a vertical asymptote at (x = 3).

Determine horizontal asymptotes: For rational functions, if the degree of the denominator is greater than the degree of the numerator, there's a horizontal asymptote at (y = 0). In this case, after simplification, the degree of the denominator (1) is greater than the degree of the numerator (which is 0 since the numerator is a constant), so there's a horizontal asymptote at (y = 0).
In summary, for the function (f(x) = \frac{x+3}{x^29}), there's a hole at ((3, \frac{1}{6})), a vertical asymptote at (x = 3), and a horizontal asymptote at (y = 0).
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To find the asymptotes or holes of ( f(x) = \frac{x + 3}{x^2  9} ):
 Factor the denominator ( x^2  9 ) to find the vertical asymptotes. ( x^2  9 ) factors as ( (x + 3)(x  3) ).
 Set the denominator equal to zero and solve for ( x ) to find vertical asymptotes. In this case, ( x + 3 = 0 ) and ( x  3 = 0 ), which gives ( x = 3 ) and ( x = 3 ) as the vertical asymptotes.
 Check for any simplifications that could lead to a hole in the graph. Simplify the function ( f(x) ) and check if any factor cancels out from both numerator and denominator.
 In this case, you can simplify ( f(x) = \frac{x + 3}{(x + 3)(x  3)} ). ( x + 3 ) cancels out, leaving ( f(x) = \frac{1}{x  3} ).
 Since the factor ( x + 3 ) canceled out, it indicates there's a hole at the point where ( x = 3 ). To find the ycoordinate of the hole, evaluate ( f(x) ) at ( x = 3 ).
 ( f(3) = \frac{1}{3  3} = \frac{1}{6} = \frac{1}{6} ). So, there's a hole at the point ( (3, \frac{1}{6}) ).
 Therefore, the vertical asymptotes are ( x = 3 ) and ( x = 3 ), and there's a hole at ( (3, \frac{1}{6}) ).
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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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