Question

Y=3+2x² знайти область значень функції

Answer 1

x²≥0 ,x∈R

2x²≥ , x∈R

3+2x²≥3 , x∈R

Otvet: y∈/3,∞)

Answer 2

У°=-Д/4а=24/4×2=3
(3;+&)

Answer 3

To find the range of the function $Y = 3 + 2x^2$, we need to determine all possible values that the function can take.

The range of a quadratic function like $Y = 3 + 2x^2$ depends on the coefficient of $x^2$ (in this case, $2$) and the constant term (in this case, $3$). Since the coefficient of $x^2$ is positive, the parabola opens upwards, and there is a minimum value, but no maximum value.

To find the minimum value of $Y$, we need to find the vertex of the parabola, which occurs at the point $(h, k)$ where $h = -\frac{b}{2a}$ and $k$ is the corresponding value of $Y$.

For the equation $Y = 3 + 2x^2$, $a = 2$ and $b = 0$ (since there is no linear term). Plugging these values into the formula for $h$, we get:

$h = -\frac{0}{2 \cdot 2} = 0$

Now, substitute $h$ back into the original function to find the value of $k$:

$k = 3 + 2 \cdot 0^2 = 3$

So, the vertex of the parabola is $(0, 3)$, and the minimum value of the function is $k = 3$. As there is no maximum value, the range of the function is $Y \geq 3$, meaning that the function can take all values greater than or equal to 3.

In interval notation, the range is $(-\infty, 3]$. This indicates that the function $Y = 3 + 2x^2$ can take any value from negative infinity up to and including 3.