Find an equivalent fraction to 5/6 only.
2 answers:
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1. Domain.
We have
in the denominator, so:
![x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\ (x-3)(x+1)\neq0\\\\\boxed{x\neq3\qquad\wedge\qquad x\neq-1}](https://tex.z-dn.net/?f=x%5E2-2x-3%5Cneq0%5C%5C%5C%5C%28x%5E2-2x%2B1%29-4%5Cneq0%5C%5C%5C%5C%28x-1%29%5E2-4%5Cneq0%5C%5C%5C%5C%28x-1%29%5E2-2%5E2%5Cneq0%5Cqquad%5Cqquad%5B%5Ctext%7Buse%20%7Da%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29%5D%5C%5C%5C%5C%28x-1-2%29%28x-1%2B2%29%5Cneq0%5C%5C%5C%5C%0A%28x-3%29%28x%2B1%29%5Cneq0%5C%5C%5C%5C%5Cboxed%7Bx%5Cneq3%5Cqquad%5Cwedge%5Cqquad%20x%5Cneq-1%7D)
So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.
2. Asymptotes:
We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)
We have
So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.
2. Asymptotes:
We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)
