Lesson 15.4: Quadratic Optimization

Property

To complete the square with multiple variables, treat each variable separately and complete the square for each one independently. For an expression like $ax^2 + by^2 + cx + dy + e$, complete the square for $x$ terms first, then for $y$ terms:

Examples

Property

A quadratic equation $y = ax^2 + bx + c$, $a \neq 0$, can be written in the vertex form

where the vertex of the graph is $(x_v, y_v)$. To convert from standard form, complete the square.

Examples

• The equation $y = 3(x - 5)^2 + 1$ is in vertex form. By comparing it to $y = a(x - x_v)^2 + y_v$, we can see the vertex is at $(5, 1)$.

• For the equation $y = -4(x + 2)^2 - 7$, we can rewrite it as $y = -4(x - (-2))^2 - 7$. The vertex is at $(-2, -7)$.

Property

For the graph of $y = ax^2 + bx + c$, the $x$-coordinate of the vertex is

To find the $y$-coordinate of the vertex, substitute the value of $x_v$ into the equation for $y$.

Examples

• To find the vertex of $y = x^2 + 8x + 10$, we identify $a=1$ and $b=8$. The x-coordinate is $x_v = \frac{-8}{2(1)} = -4$. The y-coordinate is $y_v = (-4)^2 + 8(-4) + 10 = -6$. The vertex is $(-4, -6)$.

• For the graph of $y = -2x^2 - 12x + 5$, we have $a=-2$ and $b=-12$. The x-coordinate of the vertex is $x_v = \frac{-(-12)}{2(-2)} = -3$. The y-coordinate is $y_v = -2(-3)^2 - 12(-3) + 5 = 23$. The vertex is $(-3, 23)$.