I will show you 3 methods for finding the vertex and axis of symmetry of a parabola.

Suppose an equation is as follows:

y = (x - 1)(x + 5)

First we can find the x intercepts by setting x - 1 and x + 5 equal to 0 and solve for x. When doing so, we get the intercepts to be (1, 0) and (-5, 0). The x coordinate of the vertex is halfway between 1 and -5. Therefore the x coordinate of the vertex is -2. The y coordinate of the vertex is found by substituting -2 for x in the equation.

y = (-2 - 1)(-2 + 5) = -3(3) = -9

Therefore the vertex of the parabola is (-2, -9). The axis of symmetry is the line through the x coordinate of the vertex, or x = -2 in this case.

Suppose the equation is

y = x^2 + 4x - 5.

If the equation is in the form y = ax^2 + bx + c, the x coordinate of the vertex is -b/2a.

a = 1, b = 4. Therefore the x coordinate of the vertex is -4/2(1) = -2.

The y coordinate of the vertex is y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9.

The vertex is at (-2, -9) and the axis of symmetry is x = -2.

Suppose the equation is

y + 9 = (x + 2)^2

If the equation is in the form y - k = (x - h)^2, the vertex is (h, k)

Therefore the vertex is (-2, -9) and the axis of symmetry is x = -2.

The equations y = (x -1)(x + 5), y = x^2 + 4x - 5 and y + 9 = (x + 2)^2 are the same, just written in different forms.

## No comments:

## Post a Comment