How do you write $y^{2} + 4 x + 8 y + 12 = 0$ $y^2+4x+8y+12=0$ in standard form and then graph the parabola?

Question

2704 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-22T04:37:18

Standard form of parabola = ${(y + 4)}^{2} = 4 \cdot (- 1) (x - 1)$ $(y+ 4)^2 = 4* (-1) (x-1)$

graph{-(1/4)(x^2+8x+12) [-10, 10, -5, 5]}

Axis of symmetry drawn parallel to Y-axis. May rotate it by 90^0 to make it parallel to X-axis.

Standard form of parabola is ${(y - k)}^{2} = 4 p (x - h)$ $(y - k)^2 = 4p (x - h)$

$y^{2} + 8 y + 16 = - 4 x + 4$ $y^2 +8 y + 16 = -4x +4$

${(y + 4)}^{2} = 4 \cdot (- 1) (x - 1)$ $(y+4)^2 = 4*(-1)(x - 1)$

$h = 1, k = - 4, p = - 1$ $h = 1, k = -4, p = -1$

Focus = ((h + p), k) = (0, -4) and Directrix x = h - p = 1 - (-1) = 2#

enter image source here

How do you write y2+4x+8y+12=0y^2+4x+8y+12=0 in standard form and then graph the parabola?