Answer:
[tex]\large\boxed{(2,\ 3)}[/tex]
Step-by-step explanation:
[tex]\bold{METHOD\ 1:}\\\\\text{For}\ f(x)=ax^2+bx+c\ \text{a vertex is}\ (h,\ k),\ \text{where}\ h=\dfrac{-b}{2a},\ \text{and}\ k=f(h).\\\\f(x)=x^2-4x+7\\\\a=1,\ b=-4,\ c=7\\\\\text{Substitute:}\\\\h=\dfrac{-(-4)}{2(1)}=\dfrac{4}{2}=2\\\\k=f(2)=2^2-4(2)+7=4-8+7=3\\\\\boxed{\boxed{vertex=(2,\ 3)}}[/tex]
[tex]\bold{METHOD\ 2:}\\\\\text{The vrtex form of a quadratic function:}\\\\f(x)=a(x-h)^2+k,\ \text{where}\ (h,\ k)\ \text{is a vertex}.\\\\f(x)=x^2-4x+7\qquad\text{use}\ (*)\ \ (a-b)^2=a^2-2ab+b^2\\\\f(x)=\underbrace{x^2-2(x)(2)+2^2}_{(*)}-2^2+7=(x-2)^2-4+7\\\\f(x)=(x-2)^2+3\to h=2,\ k=3\\\\\boxed{\boxed{vertex=(2,\ 3)}}[/tex]
