Find the points of discontinuity and any holes.
y=x^2-1/(x^2+3x+2)
please and thank you
y =(x^2-1)/((x+1)(x+2))
=(x-1)/(x+2)
we cancelled the (x+1)/(x+1) which is 0/0 for x=-1
so in our simplified expression when x=-1 y = -2
There is a "hole" at (-1,-2)
in the final expression the denominator cannot be zero, it is zero when x=-2
so the function is discontinuous at x=-2
so what are the points of discontinuity?
i thought that -1 and -2 were the pts of discontinuity
doesn't y=x^2-1/(x^2+3x+2)
=(x+1)(x-1)/(x-3)(x+1)
=x-1/x-3
since you factor it?
Apologies for the confusion in my previous response. Let's go through the function again and determine the points of discontinuity and any holes.
The original function is y = x^2 - 1/(x^2 + 3x + 2).
To find the points of discontinuity and any holes, we first need to simplify the function.
We can rewrite the function as y = (x^2 - 1)/((x+1)(x+2)).
To simplify further, notice that (x^2 - 1) can be factored as (x + 1)(x - 1):
y = (x + 1)(x - 1)/((x + 1)(x + 2)).
Now, we can cancel out the common factor (x + 1) in the numerator and denominator:
y = (x - 1)/(x + 2).
Now, let's analyze the simplified function:
When x = -2, the denominator becomes zero, which means the function is undefined at x = -2. Therefore, we have a vertical asymptote at x = -2, indicating a point of discontinuity.
When x = -1, we canceled out the common factor (x + 1) in the numerator and denominator. This means that the function is defined at x = -1. However, since we canceled out (x + 1), there is a "hole" in the graph at x = -1. The hole occurs at the point (-1, -2).
To summarize:
- There is a point of discontinuity at x = -2.
- There is a "hole" at x = -1, specifically at the point (-1, -2).
So, the points of discontinuity and the hole in the graph are (-1, -2) and x = -2, respectively.