Question

Help

Answer 1

@Vocaloid

Answer 2

Help with this

Answer 3

@Vocaloid

Answer 4

we only have a y^2 term not an x^2 term so start by completing the square with y^2

Answer 5

steps for completing the square (would suggest writing this down)

1. take the b-coefficient, divide it by 2, then square it, the b-coefficient being the coefficient of the term with exponent 1, so in this case, the coefficient of y

Answer 6

-8x+16
-8(x-2)
y^2+4y+4
(y+2)^2

Answer 7

getting a little ahead of yourself but you're on the right track

the y terms become (y+2)^2, since we added 4 to the left side we add 4 to the right side

so

(y+2)^2 = 8x - 4 + 4

solve for x, leave the polynomial in its factored form

Answer 8

x=y^2/8+y/2+1/2

Answer 9

you can leave the polynomial in its factored form

so it's just x = (1/8)(y+2)^2 and that's your standard form for II)

Answer 10

anyway, for part II) what kind of conic is it? given this chart:

keep in mind you only have an x term not an x^2 term

Answer 11

parabola

Answer 12

awesome

for part III) let's compare the equation

x = a(y-k)^2 + h
with our equation
x = (1/8)(y+2)^2 + 0

what are our h and k values?

Answer 13

0 and -2?

Answer 14

good so your vertex = (0,-2) and that's it for part III