Question

Find an equation in standard form of the parabola passing through the given points.
(0,3).(1,2),(2,3)

Answer 1

@jim_thompson5910

Answer 2

how far did you get?

Answer 3

I don't know where to even start.

Answer 4

what is the general equation for a parabola?

Answer 5

or a quadratic

Answer 6

Like the quadratic equation?
\[x=\frac{-b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 7

kinda, but I was aiming more towards y = ax^2 + bx + c

Answer 8

look familiar?

Answer 9

Yes sorry.

Answer 10

so we have one point (0,3) which means x = 0, y = 3

plug them into y = ax^2 + bx + c

y = ax^2 + bx + c

3 = a(0)^2 + b(0) + c

3 = a*0 + b*0 + c

3 = 0 + 0 + c

3 = c

c = 3

Answer 11

What do you get when you plug (1,2) into y = ax^2 + bx + c

Answer 12

2=a+b+c

Answer 13

good, since c = 3, we can plug that in as well

2=a+b+c

2=a+b+3

2-3=a+b

-1 = a+b

a+b = -1

Answer 14

From there, we can isolate one variable. Let's pick on b

a+b = -1

b = -1-a

b = -a - 1

Answer 15

What do you get when you plug (2,3) into y = ax^2 + bx + c

Answer 16

3=4a+2b+c
a=1

Answer 17

So
y=x^2-2x+3

Answer 18

let me check

Answer 19

3=4a+2b+c

3=4a+2b+3 ... plug in c = 3

3=4a+2(-a - 1)+3 ... plug in b = -a - 1

3=4a-2a - 2+3

3=2a + 1

3-1 = 2a

2 = 2a

2a = 2

a = 2/2

a = 1

so you are correct

Answer 20

b = -a - 1

b = -1-1

b = -2

also correct

Answer 21

so yes, y = x^2 - 2x + 3

Answer 22

I recommend checking that each point lies on this parabola.

Answer 23

Ok thanks.

Answer 24

you're welcome