Question

=

Answer 1

\(y-k)^2=4p(x-h)\) has vertex \((h,k)\) is a start

Answer 2

so (2, 6)?

Answer 3

close

Answer 4

since you have \(-4(x+2)\) it should be \((-2,6)\)

Answer 5

focus is -1?

Answer 6

the next thing we need to do is find what this looks like
since the \(y\) term is squared it opens left or right, not up or down
and since you have a \(-4\) in front of the \(x+2\) term it opens to the left

Answer 7

since \(p=4p=-4\) you know \(p=-1\)
therefore the focus is one unit TO THE LEFT of the vertex
that is why we needed the vertex first

Answer 8

notice that the focus is a coordinate (ordered pair) not a number, so i makes no sense to say that the focus is \(-1\)
what is one unit to the left of \((-2,6)\)?

Answer 9

(-1, 6)..?

Answer 10

no, that is one unit to the right

Answer 11

thats left on x axis?

Answer 12

yes, but it is to the right of \((-2,6)\)

Answer 13

start at \((-2,6)\) and go one unit to the left

Answer 14

or -3 im dumb

Answer 15

so (-3,6), makes the directrix 1?

Answer 16

no you are not dumb, and yes, it is \((-3,6)\)

Answer 17

now the directrix is a vertical line, so it looks like \(x=number\)

Answer 18

since the focus is one unit to the left, the directrix is one unit to the right of the vertex
what is one unit to the right of \((-2,6)\)?

Answer 19

(-1), but isnt the directrix negative p, if p is -1 it should be 1?

Answer 20

don't fret about the signs, they tell you which way it is open, up down or right left

Answer 21

one unit to the left is because \(p\) is negative
one unit to the right now for the directrix

Answer 22

so just find the coordinate one unit to the right of \((-2,6)\)

Answer 23

(-1,6)

Answer 24

correct?

Answer 25

yes, but don't forget the directrix is a line (unlike the focus, which is a point) so you want to write \(x=-1\) for the directrix

Answer 26

im confused because i thought it was -p, so shouldnt it be 1?