1) Why h and k not x and y? 2) If x coordinate formula is h=-b/2a then what is formula for y coordinate? 3) can you create and slove some problem to illustrate practically. 4) can you describe in more details what you mean by This formula comes from the process of finding the critical point (the vertex) of the quadratic function.

Sure, let's address each of your questions in detail:

### 1. Why \(h\) and \(k\) instead of \(x\) and \(y\)?

The notation \(h\) and \(k\) is commonly used to denote the coordinates of the vertex of a parabola in the vertex form of a quadratic function. The vertex form of a quadratic equation is:

\[ y = a(x - h)^2 + k \]

In this form, \((h, k)\) represents the vertex of the parabola. The notation \(h\) and \(k\) helps to clearly distinguish the vertex coordinates from the general \(x\) and \(y\) coordinates of any other point on the parabola.

### 2. Formula for the y-coordinate of the vertex

Once you have the x-coordinate \(h = -\frac{b}{2a}\), you can find the y-coordinate by substituting \(h\) back into the original quadratic equation \(y = ax^2 + bx + c\). The formula for the y-coordinate \(k\) of the vertex is:

\[ k = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c \]

Simplifying this, we get:

\[ k = \frac{4ac - b^2}{4a} \]

### 3. Practical problem and solution

Let's consider a quadratic function and find its vertex:

\[ f(x) = 2x^2 - 4x + 1 \]

Step 1: Find the x-coordinate of the vertex using \(h = -\frac{b}{2a}\).

Here, \(a = 2\) and \(b = -4\).

\[ h = -\frac{-4}{2 \cdot 2} = \frac{4}{4} = 1 \]

Step 2: Find the y-coordinate of the vertex by substituting \(h = 1\) back into the equation.

\[ k = 2(1)^2 - 4(1) + 1 \]
\[ k = 2 - 4 + 1 = -1 \]

So, the vertex of the parabola \(f(x) = 2x^2 - 4x + 1\) is \((1, -1)\).

### 4. Explanation of the vertex formula and critical point

The vertex formula \(-\frac{b}{2a}\) is derived from the need to find the critical point of the quadratic function, where the derivative (or slope) is zero. For a quadratic function \(f(x) = ax^2 + bx + c\), the first derivative \(f'(x)\) is:

\[ f'(x) = 2ax + b \]

Setting the derivative equal to zero to find the critical point:

\[ 2ax + b = 0 \]

Solving for \(x\):

\[ x = -\frac{b}{2a} \]

This \(x\)-value is where the function has its maximum or minimum value (the vertex). The corresponding \(y\)-value is found by substituting this \(x\)-value back into the original quadratic equation.

### Would you like more details or have any specific questions?

Here are 5 numbered questions you might want to ask:

1. Can you show another example of finding the vertex for a different quadratic function?
2. How does the sign of \(a\) affect the direction of the parabola?
3. Can you explain the process of completing the square in detail?
4. How do you determine if the vertex is a maximum or minimum point?
5. What are the applications of finding the vertex in real-world problems?

Tip: Always double-check your calculations, especially when dealing with quadratic equations, as small errors can lead to incorrect vertex coordinates.

Explore the concepts of quadratic equations, including vertex form, critical points, and practical examples. Learn how to find the vertex coordinates (h, k) using the vertex formula -b / (2a) and its application in solving quadratic functions.

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