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# Function watch

1. Need a formula for horizontal and vertical asymptote
2. 1/x.
3. 1/(y-5)
4. Ex[1] The graph ofhas a vertical asymptote at x = ________.
a. To findthe vertical asymptotes we need to set the denominator = 0 and solve.
b. Doingso gives: x2 - 1 = 0 which is (x-1)(x+1)=0. This gives thevalues of x=1 and x=-1.
c. However, x=1, is a removable discontinuity, so the answer is x = -1.
Now my question is what is the meaning of removable discontinuity
5. (Original post by shady2.0)
Ex[1] The graph ofhas a vertical asymptote at x = ________.
a. To findthe vertical asymptotes we need to set the denominator = 0 and solve.
b. Doingso gives: x2 - 1 = 0 which is (x-1)(x+1)=0. This gives thevalues of x=1 and x=-1.
c. However, x=1, is a removable discontinuity, so the answer is x = -1.
Now my question is what is the meaning of removable discontinuity
Removable in the sense that it cancels the x-1 that is also at the numerator.
6. (Original post by Onlineslayer)
Removable in the sense that it cancels the x-1 that is also at the numerator.
Thank you sir

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