How to prove that $x=b$ is the line of symmetry for the graph of $y=a(x-b)^2+c$.

Question

For $y=a(x-b)^2+c$, I know that $c$ is the extrema, and that it reaches that extrema when $x=b$; thus, the vertex of the graph is at the point $(b,c)$. I also know that the axis of symmetry is perpendicular to the line that is tangent to the vertex, thus, that line is $x=b$.

I am unsure about how to prove this, seemingly, obvious statement.

Answer 1

Hint: Prove that $f(b+k)=f(b-k)$, where $k$ is some arbitrary constant. If $f(b+k)=f(b-k)$, then the function is symmetric around the line $x=b$.

Answer 2

Hint

prove that $x\mapsto f(x+b)$ is even.