Solving for Positivity: When Does the Quadratic y = -x² + 4x - 3 Return Greater Than Zero?

To solve the problem and determine for which values of $x$ the function $y = -x^2 + 4x - 3$ is greater than 0, we proceed with the following steps:

• Step 1: Identify the roots of the quadratic equation.
• Step 2: Analyze the sign of the function in the intervals determined by the roots.

Now, let us work through each step:

Step 1: Calculate the roots using the quadratic formula. The quadratic equation is $-x^2 + 4x - 3 = 0$. Using $a = -1$, $b = 4$, $c = -3$, we apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \times (-1) \times (-3)}}{2 \times (-1)}$

$x = \frac{-4 \pm \sqrt{16 - 12}}{-2} = \frac{-4 \pm \sqrt{4}}{-2}$

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

This gives roots: $x = 1$ and $x = 3$.

Step 2: With roots at $x = 1$ and $x = 3$, the real number line is divided into intervals: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

We test a point from each interval to determine the sign of the function:

• For $x \in (-\infty, 1)$, test $x = 0$:
  $y = -(0)^2 + 4(0) - 3 = -3$ (negative).
• For $x \in (1, 3)$, test $x = 2$:
  $y = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1$ (positive).
• For $x \in (3, \infty)$, test $x = 4$:
  $y = -(4)^2 + 4(4) - 3 = -16 + 16 - 3 = -3$ (negative).

Therefore, the function is positive in the interval $(1, 3)$.

Thus, the solution is that the function $f(x) > 0$ for $1 < x < 3$.

Therefore, the correct choice is: $1 < x < 3$.