Formula to find the maximum or minimum value of a quadratic

📚 Understanding Quadratic Functions

A quadratic function is a polynomial function of degree two, generally written in the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, and $a \neq 0$. The graph of a quadratic function is a parabola.

• 🔍 The sign of $a$ determines whether the parabola opens upwards ($a > 0$, minimum value) or downwards ($a < 0$, maximum value).
• 💡 The vertex of the parabola represents the maximum or minimum point of the function.
• 📝 Finding the vertex is key to determining the maximum or minimum value.

🧭 The Vertex Formula

The vertex of a parabola given by $f(x) = ax^2 + bx + c$ can be found using the following formula for the x-coordinate ($h$):

$h = \frac{-b}{2a}$

Once you have the $x$-coordinate ($h$), you can find the $y$-coordinate ($k$) by substituting $h$ back into the original equation: $k = f(h)$. The vertex is then the point $(h, k)$. The maximum or minimum value of the quadratic function is the $y$-coordinate, $k$, of the vertex.

🧪 Completing the Square

An alternative method to find the vertex is by completing the square. This involves rewriting the quadratic function in vertex form:

$f(x) = a(x - h)^2 + k$

Where $(h, k)$ is the vertex of the parabola. To complete the square:

1. 📝 Start with $f(x) = ax^2 + bx + c$.
2. 💡 Factor out $a$ from the first two terms: $f(x) = a(x^2 + \frac{b}{a}x) + c$.
3. 🧭 Add and subtract $(\frac{b}{2a})^2$ inside the parentheses: $f(x) = a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2) + c$.
4. 🔑 Rewrite as a squared term: $f(x) = a(x + \frac{b}{2a})^2 - a(\frac{b}{2a})^2 + c$.
5. ➗ Simplify to vertex form: $f(x) = a(x - h)^2 + k$, where $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$.

📈 Real-World Examples

Example 1: Projectile Motion

A ball is thrown upwards with an initial velocity. Its height $h(t)$ at time $t$ is given by $h(t) = -5t^2 + 30t + 2$. Find the maximum height the ball reaches.

1. 🔑 Identify $a = -5$ and $b = 30$.
2. ➗ Find $t$ for the maximum height: $t = \frac{-b}{2a} = \frac{-30}{2(-5)} = 3$.
3. 💡 Find the maximum height: $h(3) = -5(3)^2 + 30(3) + 2 = -45 + 90 + 2 = 47$.

Therefore, the maximum height the ball reaches is 47 units.

Example 2: Maximizing Profit

A company's profit $P(x)$ from selling $x$ items is given by $P(x) = -0.1x^2 + 10x - 100$. Find the number of items the company should sell to maximize profit.

1. 🧭 Identify $a = -0.1$ and $b = 10$.
2. 📝 Find $x$ for maximum profit: $x = \frac{-b}{2a} = \frac{-10}{2(-0.1)} = 50$.
3. 🔍 Therefore, the company should sell 50 items to maximize profit.

✅ Practice Quiz

1. ❓ Find the minimum value of $f(x) = 2x^2 - 8x + 10$.
2. ❓ Find the maximum value of $f(x) = -x^2 + 6x - 5$.
3. ❓ A farmer wants to fence a rectangular area with 100 meters of fencing. What dimensions maximize the area? (Hint: Area = $lw$, Perimeter = $2l + 2w = 100$)

💡 Key Takeaways

• 📖 Understanding the vertex form and the vertex formula is crucial.
• ➕ Completing the square provides an alternative method to find the vertex.
• ➗ Real-world problems often involve finding maximum or minimum values of quadratic functions.