How do you identify all asymptotes or holes for #f(x)=(-x^2+5x-4)/(x^2-7x+12)#?

Question

Sophie Adams · Accepted Answer

hole at #(4,-3)#
vertical asymptote at x = 3
horizontal asymptote at y = - 1

A hole or point discontinuity is indicated when there is a common factor on the numerator/denominator of f(x). Factorising f(x) #f(x)=-(cancel((x-4))(x-1))/(cancel((x-4))(x-3))=(-(x-1))/(x-3)# Since a factor (x-4) has been removed from the denominator this indicates a discontinuity and so x ≠4 substitute x = 4 into the 'simplified' f(x) #rArrf(4)=(-3)/1=-3# Hence (4 ,-3) is a hole, that is, does not exist in this graph. The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve: #x-3=0rArrx=3" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# divide numerator/denominator by x #f(x)=((-x)/x+1/x)/(x/x-3/x)=(-1+1/x)/(1-3/x)# as #xto+-oo,f(x)to(-1+0)/(1-0)# #rArry=-1" is the asymptote"# graph{(-x+1)/(x-3) [-10, 10, -5, 5]}

Ariana Alvarez · Answer

undefined To identify all asymptotes or holes for the function $ f(x) = \frac{-x^2 + 5x - 4}{x^2 - 7x + 12} $, follow these steps:

1. Factor both the numerator and denominator if possible.
2. Cancel out any common factors to identify any potential holes.
3. Determine the vertical asymptotes by setting the denominator equal to zero and solving for $ x $.
4. Check for horizontal asymptotes by analyzing the behavior of the function as $ x $ approaches positive or negative infinity.

Let's proceed with these steps:

1. Factor the numerator and denominator:
   $ -x^2 + 5x - 4 = -(x - 4)(x - 1) $
   $ x^2 - 7x + 12 = (x - 3)(x - 4) $

2. Notice that both the numerator and denominator have a common factor of $ x - 4 $, so there might be a hole at $ x = 4 $.

3. Set the denominator equal to zero and solve for $ x $ to find the vertical asymptotes:
   $ x^2 - 7x + 12 = 0 $
   $ (x - 3)(x - 4) = 0 $
   $ x = 3 $ or $ x = 4 $

4. Analyze the behavior of the function as $ x $ approaches positive or negative infinity to find horizontal asymptotes. Since the degree of the numerator and denominator is the same (both 2), the horizontal asymptote can be determined by looking at the leading coefficients:
   $ \lim_{x \to \infty} \frac{-x^2 + 5x - 4}{x^2 - 7x + 12} = \lim_{x \to \infty} \frac{-1}{1} = -1 $
   $ \lim_{x \to -\infty} \frac{-x^2 + 5x - 4}{x^2 - 7x + 12} = \lim_{x \to -\infty} \frac{-1}{1} = -1 $

So, there is a horizontal asymptote at $ y = -1 $.

To summarize:
- There is a hole at $ x = 4 $.
- There are vertical asymptotes at $ x = 3 $ and $ x = 4 $.
- There is a horizontal asymptote at $ y = -1 $.