Y=0,5x^2-2
The equation you provided, \( Y = 0.5x^2 - 2 \), represents a quadratic function in the form of \( Y = ax^2 + bx + c \), where:
- \( a = 0.5 \)
- \( b = 0 \) (there's no linear term)
- \( c = -2 \)
Characteristics of the Quadratic Function:
Shape: Since \( a > 0 \), the parabola opens upwards.
Vertex: The vertex of a quadratic function in standard form can be found using the formula: \[ x = -\frac{b}{2a} \] Substituting \( b = 0 \) and \( a = 0.5 \): \[ x = -\frac{0}{2 \times 0.5} = 0 \] Now, substitute \( x = 0 \) back into the equation to find \( Y \): \[ Y = 0.5(0)^2 - 2 = -2 \] Therefore, the vertex is at \( (0, -2) \).
Y-intercept: The y-intercept occurs when \( x = 0 \): \[ Y = 0.5(0)^2 - 2 = -2 \] Thus, the y-intercept is at \( (0, -2) \).
X-intercepts: To find the x-intercepts, set \( Y = 0 \): \[ 0 = 0.5x^2 - 2 \] Rearranging gives: \[ 0.5x^2 = 2 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \] Therefore, the x-intercepts are at \( (2, 0) \) and \( (-2, 0) \).
Summary of Key Points:
- Vertex: \( (0, -2) \)
- Y-intercept: \( (0, -2) \)
- X-intercepts: \( (2, 0) \) and \( (-2, 0) \)
- The parabola opens upwards.
If you need further analysis or a specific type of graph for this function, let me know!
