Question

Y=2x^2-4x Найдите нули этой функции и координаты вершины ее графика.

Answer 1

Координаты вершины: х0=4/4=1, у0=2*1-4*1=-2 Нули функции: х=0, х=2

Answer 2

Finding Zeros and Vertex of the Function

The given function is Y=2x^2-4x. To find the zeros of this function and the coordinates of its vertex, we can use the standard formulas for quadratic functions.

The zeros of the function can be found by setting Y to 0 and solving for x. The formula for finding the zeros of a quadratic function is given by the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).

The vertex of the parabola represented by the function Y=2x^2-4x can be found using the formula x = -b / (2a) and then substituting the value of x into the original function to find the corresponding y-coordinate.

Let's calculate the zeros and the vertex of the function using the given formula.

The given function is Y=2x^2-4x.

The zeros of the function can be found using the quadratic formula: x = (-(-4) ± √((-4)^2 - 4*2*0)) / (2*2)

Solving for x, we get: x = (4 ± √(16)) / 4 x = (4 ± 4) / 4 x = (4 + 4) / 4 = 8 / 4 = 2 x = (4 - 4) / 4 = 0 / 4 = 0

So, the zeros of the function are x = 2 and x = 0.

Now, let's find the coordinates of the vertex of the parabola represented by the function Y=2x^2-4x.

The x-coordinate of the vertex can be found using the formula: x = -(-4) / (2*2) = 4 / 4 = 1

Substituting x = 1 into the original function, we get: Y = 2(1)^2 - 4(1) = 2 - 4 = -2

So, the coordinates of the vertex of the parabola are (1, -2).

Therefore, the zeros of the function Y=2x^2-4x are x = 2 and x = 0, and the coordinates of the vertex of its graph are (1, -2).