Answer:
Asymptotes are the points where the function tends to a given value, and the discontinuities are the points where the denominator is zero.
The function is:
[tex]f(x) = \frac{x + 1}{(x^2 + 3x + 2)}[/tex]
The discontinuitys are when x^2 + 3x + 2 = 0
So we need to find the roots of that quadratic equation, and those are:
[tex]x = \frac{-3 +-\sqrt{3^2 - 4*3*1} }{2*1} = \frac{-3+-\sqrt{-3} }{2}[/tex]
The number inside the square root, the determinant, is smaller than zero, this means that the roots are imaginary, so there is no real number x such that the denominator is equal to zero, so we do not have any discontinuity in this equation.
Now, we may have asymptotes as x goes to infinity and -infinity.
Because grade of the polynomial in the denominator is bigger than the one in the numerator, as x goes to infinity, the function will go asymptotically to +0, and as x goes to minus infinity, the function will trend asymptotically to -0
