Brainliest answer will be given
determine if the function is increasing, decreasing, even, odd, and/or invertible on its natural domain: f(x)=x^2-2x+3
state all the things that applies to it. for ex. in it's both increasing and decreasing then say both (don't leave out anything)

None

For $f(x) = x^2 - 2x + 3$, we have that $f(0) = 3$, $f(1) = 2$, and $f(2) = 3$, so $f(0) > f(1)$ and $f(1) < f(2)$. Hence, the function $f(x)$ is neither increasing nor decreasing. We have that

\[f(-x) = (-x)^2 - 2(-x) + 3 = x^2 + 2x + 3,\]

which is equal to neither $f(x)$ nor $-f(x)$ (which we can determine by plugging in, say, $x = 1$), so $f(x)$ is neither even nor odd.

From $f(0) = 3$ and $f(2) = 3$, we see that there are two different values of $x$ for which $f(x) = 3$. Therefore, $f(x)$ is not invertible

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation:

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a