Question

i need help

Answer 1

@mhanifa

Answer 2

so you already know the method

Answer 3

no

Answer 4

?

Answer 5

ok

Answer 6

ok so the form is y = a (x - h)^2 + k

Answer 7

ok

Answer 8

we substitute the values of h and k.
\[y = a(x + 4)^{2} + 2 \]

Answer 9

substitute the point (0, -30) for x and y
\[-30 = a (0 + 4)^{2} + 2 \]

Answer 10

ok

Answer 11

do we have to substitute anything else

Answer 12

no just solve for a

Answer 13

ok

Answer 14

a=-2

Answer 15

yes
ok

Answer 16

whats next

Answer 17

\[y = -2(x + 4)^{2} + 2 \]

Answer 18

what do i do with tht

Answer 19

y = 0

\[-2 x ^{2} - 16 x - 30 = 0 \]
\[-2(x ^{2} + 8 x + 15) = 0 \]
\[x ^{2} + 8 x + 15 = 0 \]
(x + 3)(x + 5) = 0


x = -3
x = -5

Answer 20

ok so what is the answer

Answer 21

Write your answer in this form: (x1, y1),(x2, y2).

Answer 22

so x1 and x2 is -3 an -5

Answer 23

this is hard what grade is this?

Answer 24

with what

Answer 25

10th

Answer 26

whats y1 and y2

Answer 27

nope

Answer 28

but uhm thank you anyways

Answer 29

why