Precalculus (10th Edition)

$C.$
If $f(x)=ax^2+bx+c$, then if $a\gt0$, the graph opens up, hence it has a maximum, if $a\lt0$, the graph opens down, hence it has a minimum. The minimum value of the graph is at its vertex, where $x=-\frac{b}{2a}$. Let's compare $f(x)=x^2-1$ to $f(x)=ax^2+bx+c$. We can see that $a=1$, $b=0$, $c=-1$. $a\gt0$, hence the graph opens up, hence it's vertex is a minimum. The minimum value is at the vertex, so: $x=-\frac{b}{2a}=-\frac{0}{2\cdot(1)}=0.$ Hence the minimum value is $f(0)=(0)^2-1=-1.$ Thus, the vertex is at $(0,-1)$ The graph that opens upward and has its vertex at $(0, -1)$ is the one in Option $C.$