What is the focus of the parabola y = –1/16 (x + 3)^2 – 6 ?

Question

amistre64 · Answer

its the point that is on the axis of symmetry a distance of "a" from the vertex :)

amistre64 · Answer

the vertex looks to be (-3,-6)

amistre64 · Answer

(y+6) = –1/16 (x + 3)^2

-16 (y+6) = (x+3)^2

relate this to:

4a y = x^2

4a = -16;  a = 4

amistre64 · Answer

F = (1,6) maybe?

amistre64 · Answer

ugh, (-3,-10) according to the wolf

agentnao · Answer

Hm, well, my notes say that the focus is (0, c) (which is 4) but the options are ...

amistre64 · Answer

agentnao · Answer

Oh. Lol, yes, (-3, -10) but where did they get the 10 from?

amistre64 · Answer

i see what I did :)

the vertex is at (-3,-6)

the focus is 4 away in the open direction;

the open direction is down

(-3,-6-4) = (-3,-10)

amistre64 · Answer

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agentnao · Answer

Ohh. Oh my gosh, that makes so much sense! And if it were a sideways parabola, then "a" would have been added to the x coordinate.

amistre64 · Answer

yep

agentnao · Answer

Thank you so much :) I'll def remember that!

amistre64 · Answer

ill prolly forget it again lol

agentnao · Answer

Haha, well, I suppose eventually I will too, but at least I'll know it for the test :)