# What are the vertex, focus and directrix of  y=4x^2 + 5x + 7 ?

Given equation:

$y = 4 {x}^{2} + 5 x + 7$

$y = 4 \left({x}^{2} + \frac{5}{4} x\right) + 7$

$y = 4 \left({x}^{2} + \frac{5}{4} x + \frac{25}{64}\right) - \frac{25}{64} + 7$

$y = 4 {\left(x + \frac{5}{8}\right)}^{2} + \frac{423}{64}$

${\left(x + \frac{5}{8}\right)}^{2} = \frac{1}{4} \left(y - \frac{423}{64}\right)$

Comparing above equation with the standard form of parabola ${X}^{2} = 4 a Y$ we get

$X = x + \frac{5}{8} , \setminus Y = y - \frac{423}{64} , a = \frac{1}{16}$

Vertex of Parabola

$X = 0 , Y = 0$

$x + \frac{5}{8} = 0 , y - \frac{423}{64} = 0$

$x = - \frac{5}{8} , y = \frac{423}{64}$

$\left(- \frac{5}{8} , \frac{423}{64}\right)$

Focus of parabola

$X = 0 , Y = a$

$x + \frac{5}{8} = 0 , y - \frac{423}{64} = \frac{1}{16}$

$x = - \frac{5}{8} , y = \frac{427}{64}$

$\left(- \frac{5}{8} , \frac{427}{64}\right)$

Directrix of parabola

$Y = - a$

$y - \frac{423}{64} = - \frac{1}{16}$

$y = \frac{419}{64}$