Marcus needs to rewrite f(x) = x2 + 6x + 4 in vertex form. his answer is f(x) = ()2 – 5.

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.



The vertex form of x^2 + 6x + 4 is (x + 3)^2 - 5.

Where the vertex is ( - 3, - 5)

The answer is orbit around nucleus