To calculate the vertex of a quadratic function in the standard form:

$f(x) = ax^2 + bx + c$

Steps:

1. Find the x-coordinate of the vertex using the formula:

   $x_{\text{vertex}} = \frac{-b}{2a}$

2. Find the y-coordinate of the vertex by substituting the x-coordinate back into the original quadratic function:

   $y_{\text{vertex}} = f\left(\frac{-b}{2a}\right) = a\left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c$

Thus, the vertex $(x_{\text{vertex}}, y_{\text{vertex}})$ is the point:

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

Alternatively, if the quadratic is given in vertex form:

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

The vertex is simply at $(h, k)$, where $h$ and $k$ are constants.

Example:

For the quadratic function:

$f(x) = 2x^2 - 4x + 1$

1. Find $x_{\text{vertex}}$:

   $x_{\text{vertex}} = \frac{-(-4)}{2(2)} = \frac{4}{4} = 1$

2. Find $y_{\text{vertex}}$ by substituting $x = 1$ into $f(x)$:

   $y_{\text{vertex}} = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1$

The vertex is $(1, -1)$.

Would you like more details or have any questions?

Related Questions:

1. How do you convert a quadratic function to vertex form?
2. What does the vertex of a parabola represent in real-world applications?
3. How can you find the axis of symmetry of a quadratic function?
4. How does the value of $a$ in the quadratic function affect the graph's shape?
5. Can the vertex of a quadratic function be outside the visible part of the graph?

Tip:

The vertex gives the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.