identify and explain all the kinds of discontinuity for y = (x + 4)/(x^2 - 16)
if x = 4 or x = -4, the denominator is 0 so the function is undefined at those points.
Sketch a graph of this equation roughly between x = -5 and x = +5 and draw your conclusions based on the graph
To identify and explain the kinds of discontinuity for the function y = (x + 4)/(x^2 - 16), we need to analyze both the vertical asymptotes and the removable discontinuities (holes) if any.
1. Vertical Asymptotes:
Vertical asymptotes occur where the denominator of a rational function becomes zero. In this case, we need to find the values of x that make the denominator, (x^2 - 16), equal to zero.
(x^2 - 16) = 0
Using the difference of squares, we can factor the denominator as follows:
(x - 4)(x + 4) = 0
This gives us two solutions:
x - 4 = 0 --> x = 4
x + 4 = 0 --> x = -4
These values, x = 4 and x = -4, are the vertical asymptotes of the function because they make the denominator equal to zero.
2. Removable Discontinuity (Hole):
To check if there is a removable discontinuity (hole) in the function, we need to determine if there are any common factors in the numerator and the denominator which can be cancelled out.
The numerator, (x + 4), does not have any factors that cancel with the denominator, (x^2 - 16), except for the factor (x + 4) itself. Therefore, there is a factor common to both the numerator and the denominator which can be canceled out.
Let's factor out the common factor, (x + 4):
y = (x + 4)/(x - 4)(x + 4)
Cancelling out the common factor:
y = 1/(x - 4)
In this new expression, there is no longer any factor that is common to both the numerator and the denominator. Thus, there are no removable discontinuities (holes) in the function.
To summarize:
The function y = (x + 4)/(x^2 - 16) has vertical asymptotes at x = 4 and x = -4. There are no removable discontinuities (holes) in the function.