Quadratic Analysis: When Is y = -2x² + 8x - 6 Negative?

To solve this problem, we'll follow these steps:

• Step 1: Find the roots of the function using the quadratic formula.
• Step 2: Determine the intervals defined by the roots and test the sign of the function on these intervals.

First, let's calculate the discriminant $b^2 - 4ac$:

$b = 8$, $a = -2$, and $c = -6$.

$\text{Discriminant} = 8^2 - 4(-2)(-6) = 64 - 48 = 16$.

With a positive discriminant, the quadratic equation has two real roots. Apply the quadratic formula:

$x = \frac{-8 \pm \sqrt{16}}{2(-2)}$.

Calculate the roots:

$x_1 = \frac{-8 + 4}{-4} = 1$

$x_2 = \frac{-8 - 4}{-4} = 3$.

Now, divide the number line into intervals based on these roots: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

Test the sign of the function $f(x) = -2x^2 + 8x - 6$ in each interval:

• For $x < 1$, choose $x = 0$:

$f(0) = -2(0)^2 + 8(0) - 6 = -6$ (negative).

• For $1 < x < 3$, choose $x = 2$:

$f(2) = -2(2)^2 + 8(2) - 6 = 2$ (positive).

• For $x > 3$, choose $x = 4$:

$f(4) = -2(4)^2 + 8(4) - 6 = -6$ (negative).

Thus, $f(x) < 0$ for $x < 1$ or $x > 3$.

The solution is $x > 3$ or $x < 1$.