Determine X for Positive Values in the Quadratic Equation y = 3x² - 6x + 4

Look at the following function:

$y=3x^2-6x+4$

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

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

To solve the problem, we'll follow these steps:

• Step 1: Solve the quadratic equation $3x^2 - 6x + 4 = 0$ using the quadratic formula.
• Step 2: Determine the intervals based on the roots and analyze the sign of the quadratic in each interval.
• Step 3: Use the results to conclude for which values of $x$ the quadratic is positive.

Step 1: The quadratic formula yields the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 3$, $b = -6$, and $c = 4$.

Calculating the discriminant:

$b^2 - 4ac = (-6)^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12$.

Since the discriminant is negative ($-12$), the quadratic equation has no real roots. This means the parabola does not intersect the x-axis, and the entire graph is above or below it.

Step 2: Since $a = 3$ (positive), the parabola opens upwards. A quadratic function with no real roots and a positive leading coefficient will be entirely above the x-axis, indicating it is always greater than zero.

Step 3: Since it is positive across all $x$-values, the solution is:

The function is positive for all $x$.