Solve the Quadratic: Discover X in y=2x^2-4x+5 Where f(x) < 0

Look at the following function:

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

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

$f(x) < 0$

To find the values of $x$ where the function $y = 2x^2 - 4x + 5$ is negative:

• Step 1: Identify the direction of the parabola:
  Since $a = 2$ is positive, the parabola opens upwards, indicating that any potential minimum will be at the vertex.

• Step 2: Find the vertex:
  The vertex is at $x = -\frac{b}{2a} = -\frac{-4}{2 \times 2} = 1$.
  Substituting $x = 1$ back into the function gives: $f(1) = 2(1)^2 - 4(1) + 5 = 3$.

• Step 3: Determine if the function can be negative:
  Since the vertex provides the minimum value of the parabola and this value is positive $f(1) = 3 > 0$, the function does not have any $x$ values for which $f(x) < 0$.

Thus, the correct answer is that the function has no negative values.