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.