Question

How do you convert the following equation from standard to vertex form by completing the square: `y=3x^2+12x+5`?

Answer 1

Please see the explanation.

Explanation:

The given equation is the equation of a parabola that opens upward (or downward). The vertex form of the equation of a parabola that opens upward (or downward) is:

`y = a(x - h)^2 + k`

where `(h, k)` is the vertex and "a" is the coefficient of the `x^2` term.

Given: `y = 3x^2 + 12x + 5`

`a = 3`, therefore, add 0 in the form `3h^2 - 3h^2` to the equation:

`y = 3x^2 + 12x + 3h^2 - 3h^2 + 5`

Factor out 3 from the first 3 terms:

`y = 3(x^2 + 4x + h^2) - 3h^2 + 5`

Set the middle term in right side of the pattern, `(x - h)^2 = x^2 - 2hx + h^2`, equal to the middle term in the equation:

`-2hx = 4x`

Solve for h:

`h = -2`

Substitute the left side of the pattern into the equation:

`y = 3(x - h)^2 - 3h^2 + 5`

Substitute -2 for h:

`y = 3(x - -2)^2 - 3(-2)^2 + 5`

Combine the constant terms:

`y = 3(x - -2)^2 - 7`

The above is the vertex form.

The vertex can be read directly from the equation; it is at:

`(-2, -7)`