## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\bf{Option \ (C)}$
1) The graph of the quadratic function $f(x) = ax^2+bx+c$ is a parabola that opens: (a) Upward when $a \gt 0$ (b) Downward when $a \lt 0$ On comparing $f(x)=x^2-1$ with $f(x) = ax^2+bx+c$, we get: $a=1, b=0,c=-1; a \gt 0$. We can see that the given function shows a graph of a parabola that opens upward. 2) The minimum value of the graph is at its vertex, where $x=−\dfrac{b}{2a}$. Thus, the given function has its vertex at $x=0$ and the minimum value is $f(0)=0^2-1=1$. So, the given function has its vertex at: $(0,−1)$ Thus, the parabola that opens upward and whose vertex is at $(0,−1)$ matches with $\bf{Option \ (C)}$.