find the minimum value of the parabola y = x^2 + 4x - 1. simplify your answer and write it as a proper…

Explanation:

Step1: Complete the square

For the quadratic function $y = x^{2}+4x - 1$, we know that for a quadratic of the form $y=ax^{2}+bx + c$ ($a = 1$, $b = 4$, $c=-1$), we use the formula $y=a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}$. [ \begin{align*} y&=x^{2}+4x - 1\ &=(x^{2}+4x)-1\ &=(x^{2}+4x + 4)-4 - 1\ &=(x + 2)^{2}-5 \end{align*} ]

Step2: Determine the minimum value

Since $(x + 2)^{2}\geq0$ for all real - valued $x$ (because the square of any real number is non - negative), when $(x + 2)^{2}=0$ (i.e., $x=-2$), the function $y$ reaches its minimum value. The minimum value of $y$ is $- 5$.

Answer:

$-5$