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

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Answer 1 · 2017-09-29T18:32:34

Given: $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

The x coordinate of the focus is:

$f_{x} = - \frac{b}{2 (a)}$ $f_x=-b/(2(a))$

The y coordinate of the focus is:

$f_{y} = a f_{x}^{2} + b f_{x} + c + \frac{1}{4 a}$ $f_y=af_x^2+bf_x+c+1/(4a)$

The equation of the directrix is:

$y = f_{y} - \frac{1}{2 a}$ $y = f_y-1/(2a)$

Given: $y = x^{2} - x + 1$ $y=x^2-x+1$

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

The x coordinate of the focus is:

$f_{x} = \frac{1}{2}$ $f_x = 1/2$

The y coordinate of the focus is:

$f_{y} = {(\frac{1}{2})}^{2} - \frac{1}{2} + 1 + \frac{1}{4}$ $f_y=(1/2)^2-1/2+1+1/4$

$f_{y} = 1$ $f_y = 1$

The focus it the point $(\frac{1}{2}, 1)$ $(1/2,1)$

The equation of the directrix is:

$y = 1 - \frac{1}{2}$ $y = 1 - 1/2$

$y = \frac{1}{2}$ $y = 1/2$

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=x2−x+1y=x^2-x+1?