Which graph represents the equation y = – (x – 1)2 + 1?

On a coordinate plane, a parabola opens up. It goes through (0, 2), has a vertex at (1, 1), and goes through (2, 2).
On a coordinate plane, a parabola opens down. It goes through (0, 0), has a vertex at (1, 1), and goes through (2, 0).
On a coordinate plane, a parabola opens down. It goes through (negative 2, 0), has a vertex at (negative 1, 1), and goes through (0, 0).
On a coordinate plane, a parabola opens up. It goes through (negative 2, 2), has a vertex at (negative 1, 1), and goes through (0, 2).

The equation given is \( y = - (x - 1)^2 + 1 \). This equation represents a parabola that opens downward because of the negative sign in front of the squared term. 

Let's analyze the characteristics:

1. **Vertex**: The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. In this equation, the vertex is at \( (1, 1) \).

2. **Opening Direction**: The parabola opens downward because the coefficient of the squared term is negative.

3. **Points on the Parabola**: To find points on the parabola, you can substitute values of \( x \) and calculate \( y \).

- At \( x = 0 \): \( y = - (0 - 1)^2 + 1 = - (1) + 1 = 0 \). So, the point \( (0, 0) \) is on the parabola.
- At \( x = 2 \): \( y = - (2 - 1)^2 + 1 = - (1) + 1 = 0 \). So, the point \( (2, 0) \) is on the parabola.

Given these characteristics, the correct description is:

**Option 2:**
- The parabola opens down.
- It goes through \( (0, 0) \), has a vertex at \( (1, 1) \), and goes through \( (2, 0) \).

This is the graph that represents the equation \( y = - (x - 1)^2 + 1 \).

Would you like more details or have any questions?

Here are 5 relative questions to consider:
1. How do you find the vertex of a parabola given in standard form?
2. What does the sign of the coefficient of the squared term tell you about the direction in which a parabola opens?
3. How can you determine the axis of symmetry for a parabola?
4. What is the effect of changing the value of \( h \) in the vertex form of a parabola?
5. How do you find the x-intercepts of a parabola?

**Tip:** Always check the direction in which the parabola opens by looking at the sign of the leading coefficient in the equation.

Explore the graph of y = - (x - 1)^2 + 1, a downward-opening parabola with its vertex at (1, 1) and passing through points (0, 0) and (2, 0). Learn how to identify the vertex, direction of opening, and points on the parabola.

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