Solving Quadratic Inequality: When is y = -x² + 6x - 8 Positive?

To solve the problem, we need to determine the intervals where the quadratic function $y = -x^2 + 6x - 8$ is greater than zero.

First, let's find the roots of the equation by setting $y = 0$:
$-x^2 + 6x - 8 = 0$.

We apply the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,
where $a = -1$, $b = 6$, and $c = -8$.

Calculating the discriminant:
$b^2 - 4ac = 6^2 - 4(-1)(-8) = 36 - 32 = 4$.

Finding the roots:
$x = \frac{-6 \pm \sqrt{4}}{2(-1)}$,
$x = \frac{-6 \pm 2}{-2}$.

This gives us two roots:
$x_1 = \frac{-6 + 2}{-2} = 2$,
$x_2 = \frac{-6 - 2}{-2} = 4$.

Now, examine the sign of the function in the intervals determined by these roots: $(-\infty, 2)$, $(2, 4)$, and $(4, \infty)$. We plug test points from each interval into the original function to determine where it is positive.

• For $x \in (-\infty, 2)$, choose $x = 0$: $f(0) = -0^2 + 6 \times 0 - 8 = -8$ (negative)
• For $x \in (2, 4)$, choose $x = 3$: $f(3) = -(3)^2 + 6 \times 3 - 8 = -9 + 18 - 8 = 1$ (positive)
• For $x \in (4, \infty)$, choose $x = 5$: $f(5) = -(5)^2 + 6 \times 5 - 8 = -25 + 30 - 8 = -3$ (negative)

The interval where $f(x) > 0$ is $(2, 4)$.

Therefore, the solution to the problem is $2 < x < 4$.