Solve y=-3x²+6x-9: Finding x-Values When Function is Positive

Look at the following function:

$y=-3x^2+6x-9$

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

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

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

• Step 1: Identify the given quadratic function and its form.
• Step 2: Determine the nature of the parabola based on the leading coefficient.
• Step 3: Calculate the discriminant to find possible roots.
• Step 4: Solve for the roots using the quadratic formula if applicable.
• Step 5: Analyze the intervals determined by the roots to assess where the function is positive.
• Step 6: Conclude whether the function is positive over any interval based on the parabola's direction.

Now, let's work through each step:

Step 1: The given function is $y = -3x^2 + 6x - 9$, which is a quadratic in the standard form $ax^2 + bx + c$, where $a = -3$, $b = 6$, and $c = -9$.

Step 2: The coefficient of $x^2$ is negative ($a = -3$), indicating the parabola opens downward.

Step 3: The discriminant $\Delta$ for the quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. Calculating this:

$\Delta = (6)^2 - 4(-3)(-9) = 36 - 108 = -72$.

Step 4: A negative discriminant ($\Delta < 0$) shows that the quadratic equation $-3x^2 + 6x - 9 = 0$ has no real roots. This means the parabola does not intersect the x-axis.

Step 5: Knowing the downward opening parabola and lack of real roots, the parabola lies entirely below the x-axis, and it never becomes positive anywhere.

Step 6: Since the function is always non-positive, we conclude that the function has no positive domain.

Therefore, the solution to the problem is The function has no positive domain.