The vertex form of a quadratic equation is expressed as
y = a(x - h)^2 + k
where
h and k are the x and y coordinates of the vertex of the curve formed.
a is the leading coefficient
The standard form of a quadratic equation is written as
y = ax^2 + bx + c
The given equation is
y = - x^2 + 12x - 4
By comparing the given equation with the standard form equation, we have
a = - 1
b = 12
c = - 4
We would find the x coordinate of the vertex by applying the formula,
x = - b/2a
By substituting the given values into this formula, we have
x = - 12/2 * - 1 = - 12/- 2 = 6
We would find the y coordinate of the vertex by substituting x = 6 into the given equation. We have
y = -(6)^2 + 12(6) - 4
y = - 36 + 72 - 4
y = 32
Thus,
x = h = 6
y = k = 32
a = - 1
By substituting these values into the vertex form equation, the equation in vertex form is
y = - (x - 6)^2 + 32
1) The method used was quite different. It is correct although there were some mistakes in the work.
2) Using a(x + d)^2 + e is like using a(x - h)^2 + k but the terms inside the parentheses should have been x - d
3) a (x+d)^2+ e cannot be called the 'vertex form' of a quadratic equation because the parenthses has x + d instead of x - d
4) The shorter way is the method that i used