Question

What are the vertical and horizontal asymptotes of y = (x + 1)/(x^2 + 3x + 2)? What are the points of discontinuity and are they removable or non-removable? What are the x- and y-intercepts? What is the domain? SHOW ALL YOUR WORK HERE OR YOU WON'T GET CREDIT!

Answer 1

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:

$f(x) = \frac{x + 1}{(x^2 + 3x + 2)}$

The discontinuitys are when x^2 + 3x + 2 = 0

So we need to find the roots of that quadratic equation, and those are:

$x = \frac{-3 +-\sqrt{3^2 - 4*3*1} }{2*1} = \frac{-3+-\sqrt{-3} }{2}$

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