randRangeNonZero(-5, 5) randRangeNonZero(-5, 5) randRangeNonZero(-5, 5) A === 1 ? "" : A === -1 ? "-" : A

Given the equation:

`\qquad y = A_DISPx^2 + -2 * A * Hx + A *H * H + K`

Find the parabola's vertex.

`(` H `,` K `)`

When the equation is rewritten in vertex form like this, the vertex is the point `(\green{h}, \blue{k})`:

`\qquad y = A(x - \green{h})^2 + \blue{k}`

We can rewrite the equation in vertex form by completing the square. First, move the constant term to the left side of the equation:

`\qquad \begin{eqnarray} y &=& A_DISPx^2 + -2 * A * Hx + A * H * H + K \\ \\ y - A * H * H + K &=& A_DISPx^2 + -2 * A * Hx \end{eqnarray}`

Next, we can factor out a `A` from the right side:

`\qquad y - A * H * H + K = A(x^2 + -2 * Hx)`

We can complete the square by taking half of the coefficient of our `x` term, squaring it, and adding it to both sides of the equation. The coefficient of our `x` term is `-2 * H`, so half of it would be `-H`, and squaring that gives us `\pink{H * H}`. Because we're adding the `H * H` inside the parentheses on the right where it's being multiplied by `A`, we need to add `\pink{A * H * H}` to the left side to make sure we're adding the same thing to both sides.

`\qquad \begin{eqnarray} y - A * H * H + K &=& A(x^2 + -2 * Hx) \\ \\ y - A * H * H + K + \pink{A * H * H} &=& A(x^2 + -2 * Hx + \pink{H * H}) \\ \\ y - K &=& A(x^2 + -2 * Hx + H * H) \end{eqnarray}`

Now we can rewrite the expression in parentheses as a squared term:

`\qquad y - K = A(x - H)^2`

Move the constant term to the right side of the equation. Now the equation is in vertex form:

`\qquad y = A(x - H)^2 + K`

Now that the equation is written in vertex form, the vertex is the point `(\green{h}, \blue{k})`:

`\qquad y = A(x - \green{h})^2 + \blue{k}`

`\qquad y = A(x - \green{(H)})^2 + \blue{(K)}`

The vertex is `(\green{H}, \blue{K})`. Be sure to pay attention to the signs when interpreting an equation in vertex form.