distinguish whether the $f(x)=-x^{2}-1$ has a minimum or maximum and graph the function. justify your answer.

Explanation:

Step1: Identify the coefficient of $x^2$

The function is $f(x)=-x^{2}-1$, and the coefficient of $x^{2}$ is $a = - 1$.

Step2: Determine if it has a min or max

Since $a=-1<0$, the parabola opens down - ward. So the function has a maximum.

Step3: Find the vertex

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$. Here $b = 0$ and $a=-1$, so $x = 0$. Substitute $x = 0$ into the function $f(x)=-x^{2}-1$, we get $f(0)=-0^{2}-1=-1$. The vertex is $(0, - 1)$.

Step4: Graph - making

We can find some additional points. When $x = 1$, $f(1)=-1^{2}-1=-2$; when $x=-1$, $f(-1)=-(-1)^{2}-1=-2$. Plot the vertex $(0, - 1)$ and the points $(1,-2)$ and $(-1,-2)$ and draw a parabola opening down - ward.

Answer:

The function $f(x)=-x^{2}-1$ has a maximum. The vertex of the parabola is $(0, - 1)$ and the parabola opens down - ward.