Question

Please help! Will medal and fan!! and testimony!
What is the equation, in standard form, of a parabola that contains the following points? Show all work.
(-2, 20), (0, -4), (4, -20)

Answer 1

@confluxepic

Answer 2

OK.

Answer 3

? @confluxepic

Answer 4

@cwrw238

Answer 5

@Zale101

Answer 6

@perl

Answer 7

@surjithayer

Answer 8

y=ax^2+bx+c

(0, -4)
-4=a(0)^2+b(0)+c
-4=c

(-2, 20)
20=a(-2)^2+b(-2)-4
20=4a-2b-4
16=4a-2b ----------------- (1)

(4, -20)
-20=a(4)^2+b(4)-4
-20=16a+4b-4
-16=16a+4b ----------------- (2)
Solve equation 1 & 2 for a and b,
and then put values of a,b,c in
y = ax^2 + bx + c

Answer 9

can you help me solve them?

Answer 10

1)a= 1/2b+4
b=2a-8

2)a=-1/4b-1
b= -4a-4

Answer 11

is that right?

Answer 12

I am really bad at this...

Answer 13

\[(16 = 4a - 2b) + (-16 = 16 a + 4b) \]
\[(2(16 = 4a - 2b) ) + ( -16 = 16 a + 4b)\]
(32 + (-16) )= (8 a + 16a) + (-4b + 4b)
16 = 32a
1/2 = a

(-2, 20)
20=1/2(-2)^2+b(-2)-4
20=2 -2b-4
16= -2 -2b
2b = -18
b = -9

y = 1/2(x^2) -9x - 4