Question

$f(x) = 3x^2 - 6x + 9$.
the given equation defines the function $f$. at what point $(x, y)$ in the $xy$-plane does the graph of $y = f(x)$ reach its minimum?
$\boldsymbol{\circ}$ $(-1, 12)$
$\boldsymbol{\circ}$ $(1, -6)$
$\boldsymbol{\circ}$ $(1, 6)$

Explanation:

Step1: Find x-coordinate of vertex

For quadratic $ax^2+bx+c$, $x=-\frac{b}{2a}$. Here $a=3$, $b=-6$, so $x=-\frac{-6}{2\times3}=1$.

Step2: Find y-coordinate of vertex

Substitute $x=1$ into $f(x)$: $f(1)=3(1)^2-6(1)+9=3-6+9=6$.

Step3: Identify minimum point

The quadratic opens upward ($a>0$), so vertex is minimum: $(1,6)$.

Answer:

(1, 6)