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

Oct 22, 2017

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

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.

Explanation:

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

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

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

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

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