http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/completing-the-square.solver
go to this address for step by step instructions on this and the answer. just plug in you a, b, and c value and your off!
The vertex form: y = a(x - h)2 + k where (h, k) is the vertex.
y = x2 + 4x + 7 add and subtract 4 to the right side to complete the square
y = x2 + 4x + 4 - 4 + 7 form the square
y = (x + 2)2 + 3 write + 2 = -(-2)
y = [x - (-2)]2 + 3
Thus, the vertex is (-2, 3)
4x-x2 = 2
6
A completely factored form is one which is composed of product of factors and can't be factorized further. Let us consider two examples: x2 - 4x + 4 is not a factored form because it can be factored as (x - 2)(x - 2). (x +1)(x2 - 4x + 4) is also not a factored form because x2 - 4x + 4 can be factored further as (x - 2)(x - 2). So, the completely factored form is (x + 1)(x - 2)(x - 2).
X2 + 4xx(x + 4)=======
The vertex form is y = (x - 4)2 + 13
The vertex of the positive parabola turns at point (-2, -11)
Y = X2 - 4X - 5set to zeroX2 - 4X - 5 = 0X2 - 4X = 5halve the linear term ( - 4 ) then square it and add that result to both sidesX2 - 4X + 4 = 5 + 4factor on the left and gather terms together on the right(X - 2)2 = 9(X - 2)2 - 9 = 0==============vertex form(2, - 9)======vertex
Example function.Y = X2 - 4X + 5set to 0X2 - 4X + 5 = 0X2 - 4X = - 5Now, halve the linear coefficient, square it and add it to both sidesX2 - 4X + 4 = - 5 + 4gather terms on the right and factor on the left(X - 2)2 = -1==============Vertex form.(2, - 1)=======Vertex.
3
Y = X2 - 4X + 12 set to 0 X2 - 4X + 12 = 0 subtract 12 from each side X2 - 4X = - 12 now, take the linear term ( - 4 ), halve it, square it and add it to both sides X2 - 4X + 4 = - 12 + 4 factor the left side, gather terms on the right (X - 2)2 = - 8 add 8 to each side (X - 2)2 + 8 = 0 ------------------------ (2, 8) ------------the vertex of this function
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
4x-x2 = 2
x2 - 4x + 3 is already in standard form.
y=-x^2+4x-3
6
-2
Easier to show you a simple example as I forget the formulaic approach. X2 + 4X - 6 = 0 add 6 to each side x2 + 4X = 6 Now, halve the linear term ( 4 ), square it and add it to both sides X2 + 4X + 4 = 6 + 4 gather the terms on the right side and factor the left side (X + 2)2 = 10 subtract 10 from each side (X + 2)2 - 10 = 0 (- 2, - 10 ) -------------------the vertex of this quadratic function