Determining x for Positive Outputs: When y=-x²+4x-5 is Greater Than Zero

Look at the following function:

$y=-x^2+4x-5$

Determine for which values of $x$ the following is true:

$f\left(x\right)>0$

To solve the problem of finding when the function $y = -x^2 + 4x - 5$ is greater than zero, follow these steps:

• Step 1: Identify the function as a downward-opening parabola because the coefficient of $x^2$ is negative.
• Step 2: Use the quadratic formula to find potential real roots. This helps determine intervals of the parabola.

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

Step 3: Plug in the coefficients for our function, $a = -1$, $b = 4$, and $c = -5$.

Calculate discriminant: $b^2 - 4ac = 4^2 - 4(-1)(-5) = 16 - 20 = -4$.

Since the discriminant is negative ($-4$), there are no real roots. This means the function never crosses the x-axis, and thus it is never positive.

Consequently, the function $y = -x^2 + 4x - 5$ has no intervals where it is positive.

Therefore, the function has no positive domain.