Question

Let f(x)= 2x^2 + 6x +3. Express f(x) in the form a(x-h)^2 + k, where a, h, and k are constants. ?

Answer 1

`y=2(x+3/2)^2-3/2`

Explanation:

The form `a(x-h)^2+k` is known as the vertex form of a quadratic. From this form we can find `bbh` the axis of symmetry, `bbk` the maximum/minimum value of the function. `bba` is the coefficient of the `bb(x^2)` term.

Start by placing a bracket around the terms containing the variable:

`y=(2x^2+6x) +3`

Next, factor out the coefficient of `x^2`

`y=2(x^2+6x) +3`

You also have to divide the coefficient of the `x` term by this value, to counteract the factored value.

`y=2(x^2+6/2x) +3`

`y=2(x^2+3x) +3`

Now add the square of half the coefficient of the `x` term inside the bracket.

`y=2(x^2+3x +(3/2)^2) +3`

You also need to subtract this outside the bracket and multiply it by the `2` we factored from `x^2`

`y=2(x^2+3x +(3/2)^2) -2*(3/2)^2 +3`

Simplify outside of the bracket:

`y=2(x^2+3x +(3/2)^2) -3/2`

Form the square of a binomial:

`y=2(x+3/2)^2-3/2`