Determine X-Values Where y = -x² + 4x - 3 Falls Below Zero

To solve the problem of determining for which values of $x$ the quadratic function $y = -x^2 + 4x - 3$ is less than zero, we should first find the roots of the equation $-x^2 + 4x - 3 = 0$.

Using the quadratic formula, where $a = -1$, $b = 4$, and $c = -3$, we have:

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

Calculating the discriminant:

$b^2 - 4ac = 4^2 - 4(-1)(-3) = 16 - 12 = 4$

Since the discriminant is positive, we will have two distinct real roots:

$x = \frac{-4 \pm \sqrt{4}}{-2} = \frac{-4 \pm 2}{-2}$

This gives us:

$x = \frac{-6}{-2} = 3$ and $x = \frac{-2}{-2} = 1$

This tells us the quadratic function crosses the x-axis at $x = 1$ and $x = 3$.

To determine the sign of the function, consider test values in the intervals determined by the roots, which are: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

• For the interval $(-\infty, 1)$, test a point such as $x = 0$: $f(0) = -0^2 + 4(0) - 3 = -3$, which is less than 0.
• For the interval $(1, 3)$, test a point such as $x = 2$: $f(2) = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1$, which is greater than 0.
• For the interval $(3, \infty)$, test a point such as $x = 4$: $f(4) = -(4)^2 + 4(4) - 3 = -16 + 16 - 3 = -3$, which is less than 0.

Therefore, the solution where $f(x) < 0$ is when the variable $x$ satisfies $x < 1$ or $x > 3$.

Hence, the values of $x$ for which $f(x) < 0$ are $x > 3$ or $x < 1$.