find the minimum value of the parabola y = x² + 4x. simplify your answer and write it as a proper fraction…

Explanation:

Step1: Complete the square

For the quadratic function $y = x^{2}+4x$, we add and subtract the square of half of the coefficient of $x$. The coefficient of $x$ is $4$, half of it is $2$ and its square is $4$. So $y=x^{2}+4x=(x + 2)^{2}-4$.

Step2: Analyze the vertex - form

The vertex - form of a parabola is $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex of the parabola. For the function $y=(x + 2)^{2}-4$, $a = 1>0$, which means the parabola opens upward. The vertex of the parabola is $(-2,-4)$.

Answer:

$-4$