What are the points of discontinuity?
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Answer:
x = 9, x = 3
Step-by-step explanation:
Given
y = [tex]\frac{x-3}{x^2-12x+27}[/tex]
The denominator of the rational function cannot be zero as this would make y undefined.
Equating the denominator to zero and solving gives the values that x cannot be, that is
x² - 12x + 27 = 0
(x - 9)(x - 3) = 0
Equate each factor to zero and solve for x
x - 9 = 0 ⇒ x = 9
x - 3 = 0 ⇒ x = 3
Thus x = 9, x = 3 are excluded values and the points of discontinuity