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

Question

3392 views around the world

You can reuse this answer
Creative Commons License

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]}

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