What are the points of discontinuity? Are they all removable? Please show your work.

y=(x-5) / x^2 - 6x +5
y = (x-5) / (x^2 - 6x + 5)
Simplifying,
y = (x-5) / (x - 5)(x - 1)
y = 1 / (x-1)
The root of the denominator is discontinuity:
x = 1 : an asymptotic discontinuity
The other roots (before simplification) is also a discontinuity:
x = 5 : a removable discontinuity
If you plot this, the graph is asymptotic to x = 1, that is, as the graph approaches to x = 1, the value of y approaches infinity. Also, the graph has a hole as point x = 5, a removable/point discontinuity.
hope this helps~ `u`
Thank you !!!
To find the points of discontinuity of the function y = (x - 5) / (x^2 - 6x + 5), we need to identify the values of x where the function is undefined or where the denominator of the fraction is equal to zero.
First, let's find the values that make the denominator zero by solving the equation x^2 - 6x + 5 = 0.
Factoring the quadratic equation, we get:
(x - 1)(x - 5) = 0
Setting each factor to zero and solving for x gives us:
x - 1 = 0 or x - 5 = 0
x = 1 or x = 5
So, the denominator of the fraction becomes zero when x = 1 and x = 5. These are the potential points of discontinuity.
Now, let's analyze if these points are removable or not. A point of discontinuity is considered removable if the function approaches a finite limit at that point.
To check if the points x = 1 and x = 5 are removable, we need to examine the function's behavior as x approaches these values.
As x approaches 1, the function becomes:
y = [(x - 5) / [(x - 1)(x - 5)]]
= 1 / (x - 1)
The function 1 / (x - 1) approaches positive or negative infinity as x approaches 1 from the left or right. Hence, the point x = 1 is a non-removable point of discontinuity.
Next, as x approaches 5, the function becomes:
y = [(x - 5) / [(x - 1)(x - 5)]]
= 1 / (x - 1)
Similarly, the function 1 / (x - 1) approaches positive or negative infinity as x approaches 5 from the left or right. Therefore, the point x = 5 is also a non-removable point of discontinuity.
In summary, the points of discontinuity for the function y = (x - 5) / (x^2 - 6x + 5) are x = 1 and x = 5, and both of these points are non-removable.
To find the points of discontinuity, we need to look for values of x that make the function undefined. In this case, the function is defined for all x, except when the denominator of the function is equal to zero.
So we need to find the values of x that make the denominator zero:
x^2 - 6x + 5 = 0
To solve this quadratic equation, we can factor it:
(x - 1)(x - 5) = 0
Setting each factor equal to zero, we get:
x - 1 = 0 --> x = 1
x - 5 = 0 --> x = 5
Therefore, the points of discontinuity occur at x = 1 and x = 5.
Now, to determine if these points are removable or not, we need to evaluate the function at these points and see if the function can be simplified to remove the discontinuity.
For x = 1:
y = (x - 5) / (x^2 - 6x + 5)
y = (1 - 5) / (1^2 - 6(1) + 5)
y = (-4) / (1 - 6 + 5)
y = (-4) / (0)
This means that the function is undefined at x = 1. Since the function cannot be simplified to eliminate the discontinuity, it is not removable.
For x = 5:
y = (x - 5) / (x^2 - 6x + 5)
y = (5 - 5) / (5^2 - 6(5) + 5)
y = (0) / (25 - 30 + 5)
y = (0) / (0)
This means that the function is undefined at x = 5. Again, the function cannot be simplified to eliminate the discontinuity, so it is not removable.
In summary, the points of discontinuity for the function y = (x - 5) / (x^2 - 6x + 5) are x = 1 and x = 5, and they are not removable.