Question

Y=2x^2-4x построить график и написать решение

Answer 1

Ответ:

ДАНО

y = 2x² - 4x

ПОСТРОИТЬ график.

1) y = 2x*(x-2) = 0

Корни - точки пересечения с осью Х - 0 и 2 - две точки

2) y(o) = 0 - точка пересечения с осью У - еще точка.

3) Производная функции - Y'(x) = 4x - 4= 4*(x-1)

Корень производной - максимум - при Х=1.

Ymin(1) = - 2.

Построение. Парабола, ветви вверх, вершина в точке А(1;-2)

Дополнительные точки.

х = -1 и у = 2 + 4 = 6

х = 3 и у = 18- 12 = 6

Рисунок к задаче в приложении.

Объяснение:

Answer 2

Task

The task is to construct the graph of the function y=2x^2-4x+1 and write the solution.

Function

The given function is: y=2x^2-4x+1

To graph this function, we first need to determine the domain and range. The domain of a quadratic function is all real numbers (-∞,∞). To determine the range, we can consider the behavior of the function as x approaches positive or negative infinity. Since the leading coefficient is positive (2x^2), the function approaches positive infinity as x approaches positive or negative infinity. Therefore, the range is (-∞,∞).

Graphing

We can sketch the graph of the function by considering the vertex and whether the parabola opens up or down. To find the vertex, we calculate the x-value by taking the negative b/2a, which is 4/4=1. Plugging this back into the original function gives the y-value of y=-1. Therefore, the vertex is (1,-1). Since the leading coefficient is positive, the parabola opens up. We can now sketch the graph of the function.

Solution

The solution includes: 1) The domain is (-∞,∞) 2) The range is (-∞,∞) 3) The vertex is (1,-1) 4) The graph of the function y=2x^2-4x+1 is a parabola that opens up with a minimum at the vertex (1,-1).