Solve the Quadratic Inequality: When Is y = -5x² - 10 Less Than Zero?

Let's solve this step-by-step.

• First, identify the key features of the parabola represented by $y = -5x^2 - 10$. It is a downward-opening parabola because the coefficient of $x^2$, which is -5, is negative.
• Since the parabola opens downwards, the vertex provides the highest point on the graph. The general formula for the vertex $x$-coordinate of the quadratic function $ax^2 + bx + c$ is $-\frac{b}{2a}$. However, as mentioned, $b = 0$, so the vertex is at $x = 0$.
• Substitute $x = 0$ back into the function to find the $y$-coordinate of the vertex:
  $y = -5(0)^2 - 10 = -10$.
• This implies that the entire parabola is situated below the y-axis since the maximum value (vertex) is -10, which is already less than zero.
• Thus, the function $f(x) = -5x^2 - 10$ is always less than zero for all $x$. This is because the vertex, the highest point, is also negative, and it only opens downward from there.

Therefore, the solution to the problem is that $f(x) < 0$ for all values of $x$.