3 years ago

1.Find the zeros of f(x) = x2 - 9

1 answer:

6 0

So for the first one it is (x-3)(x+3) this is because it is a difference of squares so you square root the 9 and the minus you do both +and - in order to keep up the -9
The zeros are:x=3 and x=-3 ( this is because you change the sign in the bracket

the others should be pretty simple to do for number 2 its a complex
the third is a simple trinomial
and fourth is simple trinomial as well

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Find the domain of f(x)=x-9/x squared-81

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$f(x)=\dfrac{x-9}{x^2-81}\\\\\text{The domain:}\\\\x^2-81\neq0\qquad\text{add 81 to both sides}\\\\x^2\neq81\to x\neq\pm\sqrt{81}\\\\x\neq-9\ \wedge\ x\neq9\\\\Answer:\ \boxed{\text{All real numbers except -9 and 9}}\to\boxed{x\in\mathbb{R}-\{-9,\ 9\}}$