y=x^2+2x-3
The equation you've provided is a quadratic function in the standard form \( y = ax^2 + bx + c \), where:
- \( a = 1 \)
- \( b = 2 \)
- \( c = -3 \)
Key Features of the Quadratic Function
Vertex: The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values: \[ x = -\frac{2}{2(1)} = -1 \] To find the \( y \)-coordinate of the vertex, substitute \( x = -1 \) back into the equation: \[ y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \] Thus, the vertex is at \( (-1, -4) \).
Axis of Symmetry: The axis of symmetry can be found with the line \( x = -1 \).
Y-intercept: The \( y \)-intercept occurs when \( x = 0 \): \[ y = (0)^2 + 2(0) - 3 = -3 \] So, the y-intercept is at \( (0, -3) \).
X-intercepts: To find the x-intercepts, we set \( y = 0 \): \[ 0 = x^2 + 2x - 3 \] Factoring gives: \[ (x + 3)(x - 1) = 0 \] This yields the x-intercepts: \[ x + 3 = 0 \Rightarrow x = -3 \\ x - 1 = 0 \Rightarrow x = 1 \] Therefore, the x-intercepts are at \( (-3, 0) \) and \( (1, 0) \).
Direction of Opening: Since \( a = 1 \) (which is positive), the parabola opens upwards.
Summary
- Vertex: \( (-1, -4) \)
- Axis of Symmetry: \( x = -1 \)
- Y-intercept: \( (0, -3) \)
- X-intercepts: \( (-3, 0) \) and \( (1, 0) \)
- Opens: Upwards
If you have any specific calculations or further questions regarding this function, feel free to ask!
