Finding Where x² - 4x + 5 > 0: Solving the Quadratic Inequality

Look at the following function:

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

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

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

The given function is $y = x^2 - 4x + 5$. To find where this function is positive, we'll first analyze the properties of this quadratic.

Let's start by completing the square. We have:

To complete the square, take the coefficient of $x$, which is $-4$, halve it to get $-2$, and then square it to get $4$. Add and subtract this inside the expression:

Now, the expression is in vertex form $y = (x-2)^2 + 1$, which indicates a parabola with a vertex at $(2, 1)$ and opens upwards. The vertex is the minimum point of the function.

Since the minimum value of $y$ is 1 (when $x = 2$), and the parabola opens upwards, the function is positive for all real $x$, because $(x-2)^2 \geq 0$ for any real number $x$, making $(x-2)^2 + 1 > 0$.

Therefore, the answer is that the function is positive for all values of $x$.

In conclusion, the correct choice is:

The function is positive for all values of $x$.