How do you find the vertical asymptotes and holes of #f(x)=(x^2+4x+3)/(x+3)#?

Question

Chloe Alexander · Accepted Answer

There are no asymptotes, only a hole at #x=-3#

Let's factorise the numerator

#x^2+4x+3=(x+1)(x+3)#

Therefore,

#f(x)=(x^2+4x+3)/(x+3)=((x+1)cancel(x+3))/cancel(x+3)#

There we have a hole at #x=-3#

graph{(x^2+4x+3)/(x+3) [-7.316, 6.73, -5.28, 1.743]}

Christian Antunez · Answer

undefined To find the vertical asymptotes and holes of the function $ f(x) = \frac{x^2 + 4x + 3}{x + 3} $:

1. Identify any values of $ x $ that would make the denominator equal to zero. These values represent potential vertical asymptotes or holes in the graph.
2. Determine if any of these values also make the numerator equal to zero. If so, there may be holes at those points. To find the $ y $-coordinate of the hole, evaluate the function at that point.
3. If there are no common factors between the numerator and denominator that cancel out, the function has a vertical asymptote at $ x = a $, where $ a $ is the value(s) that makes the denominator zero and does not make the numerator zero.
4. If there are common factors that cancel out, these points represent holes in the graph rather than vertical asymptotes. To find the $ y $-coordinate of the hole, evaluate the function at the point where the common factor was canceled out.

Apply these steps to the given function $ f(x) = \frac{x^2 + 4x + 3}{x + 3} $ to find its vertical asymptotes and holes.