Respuesta :

irspow
The discontinuity occurs when the denominator is equal to zero, as it has "infinite" slope, and thus is not a real value or point.

x^2+x-12

x^2+4x-3x-12

x(x+4)-3(x+4)

(x-3)(x+4)

So discontinuities occur when x=-4 and x=3
corncise
Firstly, factorise both the numerator and denominator to simplify it.

Let y = (x^2 + 3x - 4) / ( x^2 + x - 12)

y = (x^2 + 3x - 4) / ( x^2 + x - 12)
= (x - 1)(x + 4) / (x - 3)(x + 4)
= (x - 1) / (x - 3)

Then, by long division,

(x - 1) / (x - 3) = 1 + [2 / (x - 3)]

For vertical asymptote, when y tends to infinity, (x - 3) will tend to 0. Hence, x = 3.

For horizontal asymptote, when x tends to infinity, y = 1.

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