What are the vertex, focus and directrix of  y=x^2 – 6x + 5 ?

Dec 22, 2017

Vertex $\left(3 , - 4\right)$
Focus $\left(3 , - 3.75\right)$
Directrix $y = - 4.25$

Explanation:

Given -

$y = {x}^{2} - 6 x + 5$

Vertex

$x = \frac{- b}{2 a} = \frac{- \left(- 6\right)}{2 \times 1} = \frac{6}{2} = 3$

At $x = 3$

$y = {3}^{2} - 6 \left(3\right) + 5 = 9 - 18 + 5 = - 4$

Vertex $\left(3 , - 4\right)$

Focus and Directrix

${x}^{2} - 6 x + 5 = y$

Since the equation is going to be in the form or -

${x}^{2} = 4 a y$

In this equation $a$ is focus

the parabola is opening up.

${x}^{2} - 6 x = y - 5$

${x}^{2} - 6 x + 9 = y - 5 + 9$
${\left(x - 3\right)}^{2} = y + 4$

To find the value of $a$, we manipulate the equation as -

${\left(x - 3\right)}^{2} = 4 \times \frac{1}{4} \times \left(y + 4\right)$

$4 \times \frac{1}{4} = 1$ So the manipulation didn't affect the value $\left(y + 4\right)$

The value of $a = 0.25$

Then Focus lies 0.25 distance above vertex

Focus $\left(3 , - 3.75\right)$

Then Directrix lies 0.25 distance below vertex$\left(3 , - 4.25\right)$

Directrix $y = - 4.25$