The vertex of a quadratic function $f(x) = ax^2 + bx + c$ can be found using a specific formula derived from completing the square or using calculus.

The Vertex Formula

For the quadratic function $f(x) = ax^2 + bx + c$, the vertex $(x, y)$ has its x-coordinate given by:

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

Once you have the x-coordinate, the corresponding y-coordinate (which is the value of the function at that x) can be found by substituting $x = -\frac{b}{2a}$ back into the original function:

$y = f\left(-\frac{b}{2a}\right)$

Derivation of the Vertex Formula

1. Starting with the quadratic function: $f(x) = ax^2 + bx + c$

2. Complete the square: To convert the quadratic function into vertex form $f(x) = a(x-h)^2 + k$, where $(h, k)$ is the vertex, we first complete the square.

   • Factor out $a$ from the quadratic and linear terms: $f(x) = a\left(x^2 + \frac{b}{a}x\right) + c$

   • To complete the square, add and subtract $\left(\frac{b}{2a}\right)^2$ inside the parentheses: $f(x) = a\left[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c$

   • Simplify: $f(x) = a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c$ $f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$

   • Combine the constants outside the square: $f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)$

   This gives the vertex form: $f(x) = a(x-h)^2 + k$ where: $h = -\frac{b}{2a}, \quad k = c - \frac{b^2}{4a}$

So, the vertex $(h, k)$ is located at: $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$

Summary

• X-coordinate of the vertex: $x = -\frac{b}{2a}$
• Y-coordinate of the vertex: $y = f\left(-\frac{b}{2a}\right)$

This formula helps find the highest or lowest point on the graph of a quadratic function, depending on the sign of $a$.

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Do you want to delve deeper into this formula or any of the derivations?

Here are 5 related questions you might consider:

1. How can you use the vertex formula to determine if a quadratic function opens upwards or downwards?
2. What is the relationship between the vertex and the axis of symmetry of a parabola?
3. How would you derive the vertex form directly from the standard form of a quadratic function?
4. Can the vertex formula be applied to functions other than quadratics?
5. How does the vertex change if the quadratic function is shifted horizontally or vertically?

Tip: Understanding the vertex formula is crucial for graphing parabolas quickly, as it gives you the key point around which the parabola is symmetric.