Lesson 2: The Vertex

New Concept

Today, we're focusing on the vertex: the most important point on a parabola. You'll learn to calculate its coordinates using the formula $x_v = \frac{-b}{2a}$ and use it to find maximum or minimum values in application problems.

What’s next

Next, you’ll work through interactive examples for finding the vertex. Then, you'll test your skills on a series of practice problems.

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 $(-3, 23)$.

Property

If the equation relating two variables is quadratic, then the maximum or minimum value is easy to find: It is the value at the vertex. If the parabola opens downward, there is a maximum value at the vertex. If the parabola opens upward, there is a minimum value at thevertex.

Examples

• A company's weekly profit is modeled by $P(x) = -2x^2 + 80x - 300$, where $x$ is the price of a product. The maximum profit is at the vertex. The price that maximizes profit is $x = \frac{-80}{2(-2)} = 20$ dollars.

• The height of a diver above water is given by $h(t) = -5t^2 + 10t + 3$, where $t$ is time in seconds. The minimum height doesn't apply here, but the maximum is at $t = \frac{-10}{2(-5)} = 1$ second.

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

To find a parabola's equation from its vertex $(x_v, y_v)$ and another point $(x, y)$:

1. Substitute the vertex into the vertex form: $y = a(x - x_v)^2 + y_v$.
2. Substitute the coordinates of the other point for $x$ and $y$.
3. Solve for the value of $a$.
4. Write the final equation with the value of $a$.

Examples

• A parabola has a vertex at $(3, 4)$ and passes through $(5, 12)$. Start with $y = a(x - 3)^2 + 4$. Substitute the point: $12 = a(5 - 3)^2 + 4$, which gives $8 = 4a$, so $a=2$. The equation is $y = 2(x - 3)^2 + 4$.

• A ball's path has a vertex at $(8, 12)$ and starts at $(0, 4)$. Using $y = a(x - 8)^2 + 12$, we plug in $(0,4)$: $4 = a(0 - 8)^2 + 12$. This gives $-8 = 64a$, so $a = -\frac{1}{8}$. The equation is $y = -\frac{1}{8}(x - 8)^2 + 12$.