1) The vertex form of a quadratic function is:
y = A(x - h)^2 + k, where h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.
2) The function given is: f(x) = x^2 + 6x + 4
3) To find the vertex form of that equation you have to complete squares.
These are the steps fo complete squares:
a) You will work with the first two terms: x^2 + 6x
b) divide the coefficient of the x term by 2, form the square binomial with x and the and the half of the coefficient, and subtract the square of the half of the coefficient to keep the equality. This is how you do that:
=> 6/2 = 3
=> x^2 + 6x = (x + 3)^2 - 9
c) include the third term of the trinomial:
=> x^2 + 6x + 4= (x + 3)^2 - 9 + 4
=> x^2 + 6x + 4 = (x + 3)^2 - 5
4) The vertex form of f(x) is (x + 3)^2 - 5
5) You can prove that by expanding the square binomial to come back to the orignal form of the fucntion:
(x + 3)^2 - 5 = x^2 + 2(3)(x) + 9 - 5 = x^2 + 6x + 4, which is what you wanted to prove.
Answer:
The vertex form of x^2 + 6x + 4 is (x + 3)^2 - 5.
Where the vertex is ( - 3, - 5)
