Question

find the vertex of the parabola by completeing the square

x^2-6x+8=y

Answer 1

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

Answer 2

? i dont know how to get the answer

Answer 3

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
\[y=(x-3)^2+8-9=(x-3)^2-1\]

Answer 4

you get the \(-1\) the first method by computing \[3^2-6\times 3+8\]

Answer 5

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