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

Explanation:

Step1: Rewrite in vertex - form

The general form of a quadratic function is $y = ax^{2}+bx + c$. For $y=x^{2}+x$, where $a = 1$, $b = 1$, $c = 0$. We complete the square. $y=x^{2}+x=(x+\frac{1}{2})^{2}-\frac{1}{4}$.

Step2: Identify the vertex

The vertex - form of a parabola is $y=a(x - h)^{2}+k$, and its vertex is $(h,k)$. For $y=(x+\frac{1}{2})^{2}-\frac{1}{4}$, the vertex is $(-\frac{1}{2},-\frac{1}{4})$. Since $a = 1>0$, the parabola opens upward.

Answer:

$-\frac{1}{4}$