You can't "put aside" the 1/4!!!!!!!!!
Multiply by 4...
4y = x^2 - 2x + 5
4y - 5 = x^2 - 2x
4y - 4 = x^2 - 2x + 1
4(y - 1) = (x - 1)^2
y - 1 = (1/4)(x - 1)^2
y = (1/4)(x - 1)^2 + 1
So (h, k) = (1, 1)
Hello! The question is asking me to find the vertex, focus and directrix of the parabola, and sketch its graph.
y= (1/4) (x^2 -2x+5)
I tried putting aside the 1/4 for the time being and tried completing the square
y-5 = (x^2)-2x
y-3 = (x^2)-2x+2
y-3 = (x-1)^2
Plugging back in the 1/4 I got
y-3 = (1/4)(x-1)^2
With that, my vertex is (1, 3)
The book's answer is (1,1)
Please help me solve this problem, thanks!