Solve y=-3x²-9: Finding Values Where Function is Positive

Look at the function below:

$y=-3x^2-9$

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

$f(x) > 0$

To solve this problem, let's analyze the function $f(x) = -3x^2 - 9$ and determine when it is greater than zero.

The function is a quadratic equation of the form $ax^2 + bx + c$ where $a = -3$, $b = 0$, and $c = -9$.

Our task is to find when $-3x^2 - 9 > 0$.

Step 1: Rewrite the inequality:
$-3x^2 - 9 > 0$

Step 2: Add 9 to both sides to isolate the quadratic term:
$-3x^2 > 9$

Step 3: Divide through by -3 (note that dividing by a negative flips the inequality sign):
$x^2 < -3$

Step 4: Analyze $x^2 < -3$

A square of a real number $x^2$ is always non-negative, meaning $x^2 \geq 0$. Therefore, $x^2 < -3$ is impossible since there are no real values of $x$ such that the square of $x$ is a negative number.

Conclusion: The inequality $f(x) > 0$ has no real solutions. Therefore, no values of $x$ satisfy the inequality.

The correct answer is that no values of $x$ will make $f(x) > 0$.