To analyze the function y = \frac{(3x + 1)(x – 1)}{(x + 5)} for vertical asymptotes and holes, we need to examine both the numerator and the denominator.
Step 1: Identify the Holes
Holes occur in the graph of a function where both the numerator and the denominator are equal to zero at the same value of x. First, let’s find the roots of the numerator:
- Set the numerator to zero: (3x + 1)(x – 1) = 0
- This gives us two factors: 3x + 1 = 0 and x – 1 = 0
- Solving these equations, we find:
- x = -\frac{1}{3} (from 3x + 1)
- x = 1 (from x – 1)
Next, we check the denominator:
- Set the denominator to zero: x + 5 = 0
- Solving this gives us: x = -5
Since neither values (-1/3 and 1) derived from the numerator are the same as the denominator’s root (x = -5), there are no holes in the graph.
Step 2: Identify the Vertical Asymptotes
Vertical asymptotes occur at values of x that make the denominator equal to zero while the numerator is not equal to zero. In this case, we can determine the vertical asymptote by solving:
- Set the denominator to zero: x + 5 = 0
- This leads to x = -5
Now, we check if the numerator is not zero at this value:
- Substituting x = -5 into the numerator gives us: (3(-5) + 1)(-5 – 1) = (-15 + 1)(-6) = -14(-6) = 84
Since the numerator is non-zero at x = -5, we confirm there is a vertical asymptote at:
x = -5
Conclusion
In summary:
- No holes in the graph.
- A vertical asymptote exists at x = -5.
This understanding of vertical asymptotes and holes will aid in sketching and analyzing the behavior of the function near these critical points!