Question

Найдите значение функции y (x)=x^2-4x+12 при x=-3;1;0

Answer 1

у(х)=х²-4х+12

у(-3)= (-3)²-4* (-3) +12 =9+12+12=33

y(1)= 1²-4*1+12=9

y(0)=12

Answer 2

To find the values of the function $y(x) = x^2 - 4x + 12$ at $x = -3$, $x = 1$, and $x = 0$, simply substitute the values of $x$ into the function and perform the calculations.

1. $x = -3$: $y(-3) = (-3)^2 - 4(-3) + 12$ $y(-3) = 9 + 12 + 12$ $y(-3) = 33$

2. $x = 1$: $y(1) = (1)^2 - 4(1) + 12$ $y(1) = 1 - 4 + 12$ $y(1) = 9$

3. $x = 0$: $y(0) = (0)^2 - 4(0) + 12$ $y(0) = 0 - 0 + 12$ $y(0) = 12$

So, the values of the function $y(x)$ at $x = -3$, $x = 1$, and $x = 0$ are $y(-3) = 33$, $y(1) = 9$, and $y(0) = 12$, respectively.