How to put a quadratic into vertex form?

Steps to Solve

1. Identify the standard form of the equation

First, make sure your quadratic equation is in standard form, which is $y = ax^2 + bx + c$. For example, let’s take the equation $y = 2x^2 + 8x + 3$.

2. Factor out the coefficient of $x^2$

If $a \neq 1$, factor out the coefficient $a$ from the first two terms of the equation.

For our example: $$ y = 2(x^2 + 4x) + 3 $$

3. Complete the square

To complete the square, take half of the coefficient of $x$ (which is 4 here), square it, and add it inside the parentheses. This is $\left(\frac{4}{2}\right)^2 = 4$. But remember, since we factored out 2 earlier, we also need to subtract $2 \cdot 4$ outside to balance the equation.

So we add 4 inside the parentheses and subtract $8$ outside:

$$ y = 2(x^2 + 4x + 4 - 4) + 3 $$

4. Rewrite the expression

Now rewrite the expression inside the parentheses as a square and simplify:

$$ y = 2((x + 2)^2 - 4) + 3 $$ $$ y = 2(x + 2)^2 - 8 + 3 $$ $$ y = 2(x + 2)^2 - 5 $$

5. Identify the vertex form

Now the equation is in vertex form: $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex. For our example, this is:

$$ y = 2(x - (-2))^2 - 5 $$

So the vertex is $(-2, -5)$.