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For
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.
Answer:
Step-by-step explanation:
<-- U-substitution
<-- Replace u with tanx
Therefore,