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

Question

3044 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-03T04:28:18

Given equation:

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

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

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

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

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

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

$X=x+5/8, \ Y=y-423/64, a=1/16$

Vertex of Parabola

$X=0, Y=0$

$x+5/8=0, y-423/64=0$

$x=-5/8, y=423/64$

$(-5/8, 423/64)$

Focus of parabola

$X=0, Y=a$

$x+5/8=0, y-423/64=1/16$

$x=-5/8, y=427/64$

$(-5/8, 427/64)$

Directrix of parabola

$Y=-a$

$y-423/64=-1/16$

$y=419/64$

What are the vertex, focus and directrix of y=4x^2 + 5x + 7 y=4x2+5x+7 y=4x^2 + 5x + 7 ?