Determine X for y=x²-4x+5 When f(x) < 0

Look at the function below:

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

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

$f(x) < 0$

To solve this problem, we first consider the function $y = x^2 - 4x + 5$. This is a quadratic function, and we can analyze the parabola it represents.

The standard form of a quadratic function is $y = ax^2 + bx + c$, where in our case $a = 1$, $b = -4$, and $c = 5$.

To determine if the function has any negative values, we first find the vertex of the parabola. The vertex form of a quadratic function is determined by:

• The $x$-coordinate of the vertex is given by $x = -\frac{b}{2a}$.

Plugging in our values:

$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$.

The $y$-coordinate of the vertex can be found by substituting $x = 2$ back into the equation:

$y = (2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1$.

The vertex of the parabola is $(2, 1)$, which means the minimum point of the parabola is above the x-axis at $y = 1$.

Next, we assess whether there are any real roots by finding the discriminant $\Delta$:

$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 5 = 16 - 20 = -4$.

Since the discriminant is negative ($\Delta < 0$), this indicates the parabola does not intersect the x-axis at any real point. Therefore, it never dips below the x-axis.

Given that the vertex is above the x-axis and the discriminant is negative, the quadratic function $y = x^2 - 4x + 5$ is never negative for any real $x$.

The function has no negative values.