# What are the vertex, focus and directrix of  y=x^2-3x+4 ?

Apr 9, 2017

$\text{vertex=} \left(1.5 , 1.75\right)$
$\text{focus=} \left(1.5 , 2\right)$
"directrix: y=1.5

#### Explanation: $y = a {\left(x - h\right)}^{2} + k \text{ the vertex form of parabola}$

$\text{vertex=} \left(h , k\right)$

$\text{focus=} \left(h , k + \frac{1}{4 a}\right)$

$y = {x}^{2} - 3 x + 4 \text{your parabola equation}$

$y = {x}^{2} - 3 x \textcolor{red}{+ \frac{9}{4} - \frac{9}{4}} + 4$

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

$y = {\left(x - \frac{3}{2}\right)}^{2} + \frac{7}{4}$

$\text{vertex} = \left(h , k\right) = \left(\frac{3}{2} , \frac{7}{4}\right)$

$\text{vertex=} \left(1.5 , 1.75\right)$

$\text{focus=} \left(h , k + \frac{1}{4 a}\right)$

$\text{focus=} \left(1.5 , \frac{7}{4} + \frac{1}{4 \cdot 1}\right) = \left(1.5 , \frac{8}{4}\right)$

$\text{focus=} \left(1.5 , 2\right)$

$\text{Find directrix :}$

$\text{take a point(x,y) on parabola}$

$\text{let } x = 0$

$y = {0}^{2} - 3 \cdot 0 + 4$

$y = 4$

$C = \left(0 , 4\right)$

$\text{find distance to focus}$

$j = \sqrt{{\left(1.5 - 0\right)}^{2} + {\left(2 - 4\right)}^{2}}$

$j = \sqrt{2.25 + 4}$

$j = \sqrt{6.25}$

$j = 2.5$

$\mathrm{di} r e c t r i x = 4 - 2.5 = 1.5$

$y = 1.5$