Find the minimum value of the parabola y=x^2+4x-5 Simplify your answer and write it as a proper fraction, improper fraction, or integer. square

To find the minimum value of the parabola $y=x^{2}+4x-5$, we can complete the square to rewrite the quadratic function in vertex form.<br /><br />Step 1: Start with the given equation:<br />$y = x^{2} + 4x - 5$<br /><br />Step 2: Complete the square by adding and subtracting $(\frac{4}{2})^2 = 4$ inside the parentheses:<br />$y = (x^{2} + 4x + 4) - 4 - 5$<br />$y = (x + 2)^{2} - 9$<br /><br />The vertex form of the quadratic function is $y = a(x - h)^{2} + k$, where $(h, k)$ represents the vertex of the parabola. In this case, the vertex is $(-2, -9)$.<br /><br />Since the coefficient of $(x + 2)^{2}$ is positive, the parabola opens upwards, and the vertex represents the minimum point.<br /><br />Therefore, the minimum value of the parabola $y = x^{2} + 4x - 5$ is $-9$.