Answer
See explanation
Work Step by Step
The quadratic function in standard form is:
$$f(x)=x^2-6x+2$$
We bring the equation at the vertex form:
$$f(x)=x^2-6x+2=(x^2-6x+9)-9+2=(x-3)^2-7$$
The vertex is: $$(3,-7)$$
The axis of symmetry is: $$x=3$$
We determine the $x$-intercepts: $$\begin{aligned}
(x-3)^2-7&=0\\
(x-3)^2&=7\\
x-3&=\sqrt 7\\
x&=3\pm\sqrt 7.
\end{aligned}$$
So the $x$-intercepts are:$$(3-\sqrt 7,0),(3+\sqrt 7,0)$$
We sketch the graph.
