Solve y=-x²-2x-3: Finding Where Function Values Are Negative

Look at the following function:

$y=-x^2-2x-3$

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

$f(x) < 0$

To determine when the function $f(x) = -x^2 - 2x - 3$ is negative, we begin by stating that it is a quadratic function in the form $ax^2 + bx + c$ where $a = -1$, $b = -2$, and $c = -3$. Since $a < 0$, the parabola opens downwards, indicating that it is concave down. This means that the function will be negative at all points unless it touches or crosses the x-axis.

First, we need to determine the vertex of the quadratic to ascertain where the maximum occurs. For any quadratic function in the form $ax^2 + bx + c$, the x-coordinate of the vertex is given by the formula:

$x = -\frac{b}{2a}$

Substitute $b = -2$ and $a = -1$ into the formula:

$x = -\frac{-2}{2(-1)} = -\frac{2}{-2} = 1$

The x-coordinate of the vertex is $x = 1$. The vertex lies at $(1, f(1))$.

Substitute $x = 1$ into the equation to find $f(1)$:

$f(1) = -(1)^2 - 2(1) - 3 = -1 - 2 - 3 = -6$

The vertex of the parabola is at $(1, -6)$, showing that the maximum point is negative.

Since the parabola opens downwards, all other values are below this vertex, hence **the parabola never crosses or touches the $x$-axis** implying the function is always below the $x$-axis, confirming that the function is negative for all values of $x$.

Therefore, the function is negative for all $x$.