How do you find the points of discontinuity for #y= (x-8)/(x^2+5x-6)#?

Question

Lily Anderson · Accepted Answer

By taking apart our equation, we can spot vertical asymptotes at #x=1# and #x=-6# and a horizontal asymptote at #y=0#.

In order to find our points of discontinuity, we need to factor the polynomial in the bottom. When we get the two binomials, we'll set each of them equal to 0. That way, we can find what values will create a 0 in the denominator. #y=(x+8)/((x+6)(x-1))# #x+6=0# #x=-6# #x-1=0# #x=1# You can't divide by 0, so discontinuities will be exist as vertical asymptotes at those points. So, vertical asymptotes will exist as #x=-6# and at #x=1#. Next, we can find a horizontal asymptote. The easiest way to locate them is by looking at the leading terms in both halves of the fraction. If the leading term in the denominator has a higher power than the leading therm in the numerator, then a horizontal asymptote exists at #y=0#. This applies to our equation! No holes exist in our equation because it is in its simplest form and nothing else can be taken out.

Xavier Alonso · Answer

undefined To find the points of discontinuity for $ y = \frac{{x - 8}}{{x^2 + 5x - 6}} $, follow these steps:

1. Identify any values of $ x $ that make the denominator equal to zero, as division by zero results in a discontinuity.
2. Solve the equation $ x^2 + 5x - 6 = 0 $ to find the values of $ x $ that make the denominator zero.
3. These values of $ x $ represent potential points of discontinuity for the function.

Let's solve the equation $ x^2 + 5x - 6 = 0 $ to find the values of $ x $:

$$ x^2 + 5x - 6 = 0 $$

This is a quadratic equation that can be factored as:

$$ (x - 1)(x + 6) = 0 $$

Setting each factor equal to zero gives us the solutions:

$$ x - 1 = 0 \implies x = 1 $$
$$ x + 6 = 0 \implies x = -6 $$

So, the points of discontinuity for the function $ y = \frac{{x - 8}}{{x^2 + 5x - 6}} $ occur at $ x = 1 $ and $ x = -6 $.