What are the coordinates of the vertex of the parabola with the equation \( y=x^{2}+2 x-3 \) ?

To find the vertex of the parabola given by the equation \( y = x^2 + 2x - 3 \), we can use the vertex formula \( x = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). Here, \( a = 1 \) and \( b = 2 \). Plugging these values in, we get \( x = -\frac{2}{2 \cdot 1} = -1 \). Now, to find the \( y \)-coordinate, we substitute \( x = -1 \) back into the equation: \[ y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4. \] So, the vertex of the parabola is at the coordinates \( (-1, -4) \). The vertex’s x-coordinate simplifies finding the minimum or maximum of the parabola. Here, since it opens upwards (as the coefficient of \( x^2 \) is positive), the vertex at \( (-1, -4) \) gives us the minimum point of the quadratic function—kind of like the lowest point in a valley! So, if you're plotting, that little point is your perfect starting place to understand the shape of the parabola.