find the vertex of the parabola by completeing the square

x^2-6x+8=y

Question

find the vertex of the parabola by completeing the square

x^2-6x+8=y

anonymous · Answer

first coordinate of the vertex is $-\frac{b}{2a}=-\frac{-6}{2\times 1}=3$ so it will look like 
$$y=(x-3)^2+k$$ to find $k$ replace $x$ by 3 in the expression $x^2-6x+8$ and see what you get

anonymous · Answer

? i dont know how to get the answer

anonymous · Answer

notice i did not "complete the square" but  you can do that too, it is just more annoying
start with 
$$y=x^2-6x+9$$ then think "half of 6 is 3, and $3^2=9$ and write 
$$y=(x-3)^2+8-9=(x-3)^2-1$$

anonymous · Answer

you get the $-1$ the first method by computing $$3^2-6\times 3+8$$

anonymous · Answer

b2=4ac
y=82/4(x2-6x