Final answer:
The function f(x)=1/(3x²+3x-18) has discontinuities at x=-3 and x=2, which are both vertical asymptotes since the numerator does not cancel any factors with the denominator.
Step-by-step explanation:
To identify the points of discontinuity, such as holes and vertical asymptotes, in the function f(x) = 1/(3x²+3x-18), we need to factor the denominator, if possible, and determine if there are any values for x that will make the denominator equal to zero. These values would be the ones where the function is not defined, possibly leading to vertical asymptotes or holes.
First, we can simplify the denominator by factoring out 3 from each term in the quadratic expression:
3x² + 3x - 18 = 3(x² + x - 6)
Next, we can factor the quadratic:
x² + x - 6 = (x + 3)(x - 2)
So, our factored form is:
f(x) = 1/3(x + 3)(x - 2)
Now, we identify values of x for which the denominator becomes zero:
x + 3 = 0 → x = -3
x - 2 = 0 → x = 2
Therefore, the function f(x) has discontinuities at x = -3 and x = 2. To determine if these are holes or vertical asymptotes, we examine the numerator. Given that the numerator is just 1 and does not contain factors that cancel out with the denominator, both discontinuities will be vertical asymptotes.