## Precalculus (10th Edition)

$H$
RECALL: (1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens: (i) upward when $a \gt0$ and has its vertex as its minimum; or (ii) downward when $a\lt0$ and has its vertex as its maximum. (2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$ Let's compare $f(x)=x^2-2x$ to $f(x)=ax^2+bx+c$. We can see that $a=1$, $b=-2$, $c=0$. $a\gt0$, hence the graph opens up, hence it's vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-2}{2\cdot(1)}=1.$ Hence the minimum value is $f(1)=(1)^2-2(1)=-1.$ Thus, the vertex is at $(1,-1)$ and the graph opens up. Therefore, the graph of the given function is the one in graph $H$.