How do you find the focus, directrix and sketch #y=x^2-x+1#?

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Answer 1 · 2017-09-29T18:32:34

Given: #y=ax^2+bx+c#

The x coordinate of the focus is:

#f_x=-b/(2(a))#

The y coordinate of the focus is:

#f_y=af_x^2+bf_x+c+1/(4a)#

The equation of the directrix is:

#y = f_y-1/(2a)#

Given: #y=x^2-x+1#

then #a = 1, b = -1 and c = 1#

The x coordinate of the focus is:

#f_x = 1/2#

The y coordinate of the focus is:

#f_y=(1/2)^2-1/2+1+1/4#

#f_y = 1#

The focus it the point #(1/2,1)#

The equation of the directrix is:

#y = 1 - 1/2#

#y = 1/2#

Here is a graph

graph{y=x^2-x+1 [-9.905, 10.095, -2.72, 7.28]}