Quadratic Equation Challenge: Analyze y = -x² + 5x + 6 for Positive and Negative Domains

To determine the positive and negative domains of the function $y = -x^2 + 5x + 6$, we first find the roots of the equation by solving:

$-x^2 + 5x + 6 = 0$.

We use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = -1$, $b = 5$, and $c = 6$. Substituting these values, we find:

$x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(6)}}{2(-1)} = \frac{-5 \pm \sqrt{25 + 24}}{-2} = \frac{-5 \pm \sqrt{49}}{-2}$.

$x = \frac{-5 \pm 7}{-2}$.

Solving the two scenarios regarding the $\pm$ gives $x = \frac{2}{-2} = -1$ and $x = \frac{-12}{-2} = 6$.

This means the roots are $x = -1$ and $x = 6$.

We now test the intervals defined by these roots: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.

- For $x < -1$: pick $x = -2$. Substitute into the function:

$y = -(-2)^2 + 5(-2) + 6 = -4 - 10 + 6 = -8$ (negative).

- For $-1 < x < 6$: pick $x = 0$. Substitute:

$y = -(0)^2 + 5(0) + 6 = 6$ (positive).

- For $x > 6$: pick $x = 7$. Substitute:

$y = -(7)^2 + 5(7) + 6 = -49 + 35 + 6 = -8$ (negative).

Thus, the function is positive in the interval $-1 < x < 6$ and negative in the intervals $x < -1$ and $x > 6$.

Therefore, the solution to the problem is:

$x > 0 : -1 < x < 6$

$x > 6$ or $x < 0 : x < -1$