SOLUTION: Can you please help. (33) Find the vertex, focus, and directrix of the graph of the equation 8(y+2) =(x+6)^2 (A) vertex (-6,-2) focus (-6,0) directrix is y=-4 (B) vertex (6,

Question 30594: Can you please help.
(33)
Find the vertex, focus, and directrix of the graph of the equation 8(y+2) =(x+6)^2
(A) vertex (-6,-2) focus (-6,0) directrix is y=-4
(B) vertex (6, -2) focus (-6, 4) directrix is y=-8
(C) vertex (6, -2) focus (6,0) directrix is y=4
(D) vertex (-6, -2) focus (6,4) directrix is y=o
thanks for your help
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
8(y+2) =(x+6)^2 your equation
8y + 16 = (x+6)^2 distribute the 8
8y + 16 = (x+6)(x+6) fill in: (x+6)^2 is the same as (x+6)(x+6)
8y + 16 = x^2 + 12x + 36 FOIL
8y = x^2 + 12x + 20 subtract 16
y = (1/8)x^2 + (3/2)x + 5/2
the formula for vertex is (-b/2a, f(x)) in (ax^2 + bx + c = 0)
(-b/2a)
(-3/2)/(1/4)
(-3/2)(4)
-6
f(x) is the y-value
(1/8)x^2 + (3/2)x + 5/2 plug in the -6
(1/8)(36) + (3/2)(-6) + 5/2
9/2 - 9 + 5/2
9/2 - 18/2 + 5/2
-4/2
-2
the vertex is (-6, -2)
the equation for finding foci and directrix is: p=(1/4a)
p = (1/4a)
p = 1/(1/2)
p = 2
since it is positive, the foci will be 2 units above the vertex
the foci should be at (-6, 0)
the directrix will obviously be below, so you go two units down
a point on the directrix should be at (-6, -4), but the directrix is a line: y = -4
(A) should be the correct answer
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