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

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!