Question

Графіком квадратичної функції є парабола, що має вершину у початку координат і проходить через точку А(2;-8). Задайте цю функцію формулою.

Answer 1

Графіком квадратичної функції є парабола, що має вершину у початку координат і проходить через точку А(2;-8). Задайте цю функцію формулою.

Графиком квадратичной функции является парабола, что вершину в начале координат и проходит через точку А (2; -8). Задайте эту функцию формулой
  
   Решение:
Уравнение параболы задается уравнением
y =ax²+bx+с  или х = ay²+by+с(данное уравнение можете не рассматривать)
где а≠0
Так как вершина параболы находится в начале координат то b=c=0
Уравнение параболы можно записать как:
            y =ax²  или  х = ay²(данное уравнение можете не рассматривать)
Найдем постоянную величину а из уравнений подставив координаты точки А(2;-8)
а = у/х² = -8/2² =-8/4=-2
y = -2x²
a = x/y² =2/(-8)² =2/64 =1/32
x = y²/32(данное уравнение можете не рассматривать)
Рішення :
Рівняння параболи задається рівнянням
y = ax ² + bx + з або х = ay ² + by + з
де а ≠ 0
Так як вершина параболи знаходиться на початку координат то b = c = 0
рівняння можна записати як
                                     y = ax ² або х = ay ²
Знайдемо постійну величину а з рівнянь підставивши координати точки А (2; -8)
а = у / х ² = -8 / 2 ² = -8/4 = -2
y =-2x ²
a = x / y ² = 2 / (-8) ² = 2/64 = 1/32
x = y ² / 32

Answer 2

Quadratic Function with Vertex at the Origin and Passing through Point A(2, -8)

A quadratic function is represented by the equation y = ax^2 + bx + c, where a, b, and c are constants. To find the equation of a quadratic function with a vertex at the origin (0, 0) and passing through point A(2, -8), we can substitute the coordinates of point A into the equation and solve for the constants.

Let's start by substituting the coordinates of point A into the equation:

- x = 2 - y = -8

Substituting these values into the equation, we get:

-8 = a(2)^2 + b(2) + c

Simplifying the equation, we have:

-8 = 4a + 2b + c Since the vertex of the parabola is at the origin (0, 0), we know that the value of c is 0. Therefore, we can rewrite equation as:

-8 = 4a + 2b

Now, we have one equation with two variables. To solve for a and b, we need another equation. We can use the fact that the vertex is at the origin to find another point on the parabola. Since the vertex is at (0, 0), we can substitute these values into the equation:

0 = a(0)^2 + b(0) + c

Simplifying, we get:

0 = 0 + 0 + c

Therefore, c = 0.

Now, we have two equations:

1. -8 = 4a + 2b 2. c = 0

Since c = 0, we can rewrite equation 1 as:

-8 = 4a + 2b

Now, we have one equation with two variables. We need another equation to solve for a and b. We can use the fact that the parabola passes through point A(2, -8). Substituting these values into the equation, we get:

-8 = 4a(2)^2 + 2b(2)

Simplifying, we have:

-8 = 16a + 4b

Now, we have two equations:

1. -8 = 4a + 2b 2. -8 = 16a + 4b

We can solve this system of equations to find the values of a and b.

Subtracting equation 1 from equation 2, we get:

0 = 12a + 2b

Simplifying, we have:

0 = 6a + b

Now, we have a new equation:

3. 0 = 6a + b

We can substitute equation 3 into equation 1 to solve for a:

-8 = 4a + 2(6a + b)

Simplifying, we have:

-8 = 4a + 12a + 2b

Combining like terms, we get:

-8 = 16a + 2b

Rearranging the equation, we have:

16a + 2b = -8

Now, we can substitute equation 3 into this equation:

16a + 2(0) = -8

Simplifying, we have:

16a = -8

Dividing both sides by 16, we get:

a = -0.5

Substituting this value of a into equation 3, we can solve for b:

0 = 6(-0.5) + b

Simplifying, we have:

0 = -3 + b

Adding 3 to both sides, we get:

b = 3

Therefore, the equation of the quadratic function with a vertex at the origin and passing through point A(2, -8) is:

y = -0.5x^2 + 3x

This equation represents a parabola that opens downwards and has its vertex at the origin (0, 0), passing through point A(2, -8).

Please note that the equation provided is derived based on the given information and calculations.