Question

График функции у = ах2 + bx + с проходит через точки (3; -7) и (-3; 17). Запиши уравнение функций y = ax² + bx + c, если известно, что график функции получен путем преобразования параболы у= y= x + 2 1 3 Помогите пожалуйста

Answer 1

Ответ:

y=1/3x²-4x+2

Объяснение:

Answer 2

Finding the Equation of a Parabola

To find the equation of a parabola given two points it passes through, we can use the general form of a quadratic equation: y = ax² + bx + c. In this case, we are given the points (3, -7) and (-3, 17) that the parabola passes through.

To find the values of a, b, and c, we can substitute the coordinates of the points into the equation and solve the resulting system of equations.

Let's start by substituting the coordinates of the first point (3, -7) into the equation:

-7 = a(3)² + b(3) + c

Next, we substitute the coordinates of the second point (-3, 17) into the equation:

17 = a(-3)² + b(-3) + c

Now we have a system of two equations with three unknowns (a, b, and c). To solve this system, we can use any method such as substitution or elimination. Let's use the elimination method.

Subtracting the second equation from the first equation, we get:

-7 - 17 = a(3)² + b(3) + c - (a(-3)² + b(-3) + c)

Simplifying, we have:

-24 = a(3)² + b(3) - a(-3)² - b(-3)

Simplifying further, we get:

-24 = 9a + 3b - 9a + 3b

The 9a and -9a terms cancel out, and the 3b and 3b terms cancel out as well. This leaves us with:

-24 = 0

This equation is not possible, which means there is no solution for the system of equations.

Therefore, there is no quadratic equation that satisfies the given conditions of passing through the points (3, -7) and (-3, 17) and being obtained by transforming the parabola y = x² + x + 3.

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