Graph Parabolas with Vertices Not at the Origin
-$( \mathrm { x } - 1 ) ^ { 2 } = - 12 ( \mathrm { y } + 2 )$
A) vertex: $( 1 , - 2 )$
focus: $( 1 , - 5 )$
directrix: $y = 1$
B) vertex: $( - 1,2 )$
focus: $( - 1 , - 1 )$
directrix: $y = 5$
C) vertex: $( - 2,1 )$
focus: $( - 2 , - 2 )$
directrix: $y = 4$
D) vertex: $( 1 , - 2 )$
focus: $( 1,1 )$
directrix: $x = - 5$

Question

Quizplus · Accepted Answer

The equation $(x - 1)^2 = -12(y + 2)$ is in the form $(x - h)^2 = 4p(y - k)$, where $(h, k)$ is the vertex of the parabola. Here, $h = 1$ and $k = -2$, so the vertex is $(1, -2)$. The coefficient of $y + 2$ is $-12$, indicating a vertical parabola opening downwards with $4p = -12$, so $p = -3$. The focus is $p$ units from the vertex along the axis of symmetry, so it is $(1, -2 - 3) = (1, -5)$. The directrix is a horizontal line $p$ units in the opposite direction from the focus, so it is $y = -2 + 3 = 1$.

Graph Parabolas with Vertices Not at the Origin - (x−1)2=−12(y+2)( \mathrm { x } - 1 ) ^ { 2 } = - 12 ( \mathrm { y } + 2 )(x−1)2=−12(y+2)

Graph Parabolas with Vertices Not at the Origin
- $( \mathrm { x } - 1 ) ^ { 2 } = - 12 ( \mathrm { y } + 2 )$