Y=-x^2+2x

The equation \( y = -x^2 + 2x \) represents a quadratic function. Here's a detailed analysis of the function:

1. Standard Form

The given equation can be rewritten in standard form as: \[ y = -1(x^2 - 2x) \]

2. Finding the Vertex

The vertex of a quadratic function \( y = ax^2 + bx + c \) can be found using the formula: \[ x_{\text{vertex}} = -\frac{b}{2a} \] For your equation:

• \( a = -1 \)
• \( b = 2 \)

Calculating the x-coordinate of the vertex: \[ x_{\text{vertex}} = -\frac{2}{2 \cdot -1} = 1 \]

Now, substituting \( x = 1 \) back into the equation to find the y-coordinate: \[ y_{\text{vertex}} = -1^2 + 2(1) = -1 + 2 = 1 \] So, the vertex of the parabola is at \( (1, 1) \).

3. Axis of Symmetry

The axis of symmetry of the parabola is the vertical line that passes through the vertex: \[ x = 1 \]

4. Direction of Opening

Since the coefficient of \( x^2 \) (which is \( -1 \)) is negative, the parabola opens downwards.

5. Y-Intercept

To find the y-intercept, substitute \( x = 0 \): \[ y = -0^2 + 2(0) = 0 \] Thus, the y-intercept is at the point \( (0, 0) \).

6. X-Intercepts

To find the x-intercepts, set \( y = 0 \): \[ 0 = -x^2 + 2x \] Factoring: \[ 0 = x(-x + 2) \] This gives us: \[ x = 0 \quad \text{or} \quad x = 2 \] Thus, the x-intercepts are at \( (0, 0) \) and \( (2, 0) \).

Summary:

• Vertex: \( (1, 1) \)
• Axis of symmetry: \( x = 1 \)
• Opens: Downward
• Y-intercept: \( (0, 0) \)
• X-intercepts: \( (0, 0) \) and \( (2, 0) \)

If you need further analysis or graphing tips, feel free to ask!