y=6-3x/(x- 2)(x- 3)
Recall that division by zero is not defined. Therefore, the rational function is undefined where x- 2=0 and where x- 3=0.
ccc
x- 2=0 & and & x- 3=0
⇕ & & ⇕
x= 2 & and & x= 3
This means that neither x= 2 nor x= 3 are included in the domain.
Domain
All real numbers except x= 2 and x= 3
Points of Discontinuity
If a real number a is not in the domain of a function, then the function has a point of discontinuity at x=a. Since the domain is all real numbers except x= 2 and x= 3, we can write the points of discontinuity of our function.
Points of Discontinuity
x= 2 and x= 3
To determine whether the discontinuities are removable or non-removable, we will factor the numerator and cancel out any common factors between the numerator and denominator.
Now that we have simplified the equation, let's consider the points of discontinuity, x= 2 and x= 3.
The factor x- 2 is not in the denominator any more, so x= 2 is a removable point of discontinuity.
The factor x- 3 is still in the denominator, so x= 3 is a non-removable point of discontinuity.
Intercepts
The intercepts of a function are the points where the graph intersects the axes. Let's calculate the intercepts of the given function, if there are any.
x-intercepts
The x-intercepts of a function are the points where the graph intersects the x-axis. At these points, the value of the y-coordinate is zero. Therefore, to find the x-intercepts, we have to substitute 0 for y in the function rule and solve for x.
We found that if y=0, the value of x is 2. However, x=2 is not included in the domain of the function. Therefore, there are no x-intercepts.
y-intercept
The y-intercept of a function is the point where the graph intersects the y-axis. At this point, the value of the x-coordinate is zero.
Therefore, to find the y-intercept, we have to substitute 0 for x in the function rule and solve for y.
