Question

Find the line of symmetry of the parabola defined by 3x^2-6x+8. I will give a medal

Answer 1

Give me a sec to write it on paper

Answer 2

y = -3x^2 -6x -8
.
The axis of symmetry is defined by x = -b/2a.
.
-b/2a = -(-6)/(2*(-3) = 6/-6 = -1
.
Then substitute x=-6 to find 'y'.
.


y = -3(-1^2) -(6-1) -8
y = -3 +6 -8
y = -5
.
The vertex is: (-1,-5).

Answer 3

Thank you so much!! @grrrxxiii

Answer 4

I hope that is correct. Please let me know so I can do some research if I am wrong.

Answer 5

no what he said is wrong ok so the derivative is 6x-6 which is =0 so x=1 Any y=17

Answer 6

sorry y=5***

Answer 7

The answer my teacher gave me was x=1. that is all she said

Answer 8

Can you show me the steps? I think that is where i am confused @joyraheb

Answer 9

have you took derivative?

Answer 10

I dont know what that is

Answer 11

ok forget so it is defined by x=-2b/a

Answer 12

i mean -b/2a

Answer 13

Okay whats next?

Answer 14

amd a=6 and b=-3 so you get 6/-(-2*3) which is 1

Answer 15

thats x so then you plug 1 for x in the equ to find y

Answer 16

Oh ok!! Thank you soooooo much! i still have more if you're available to help?

Answer 17

so 3-6+8=5 so y=5 so the coord are (1,5)

Answer 18

ye no prob

Answer 19

What about what is the factored form of the equation 3x^2+6-10

Answer 20

you mean 6x-10?

Answer 21

oops i meant x not 6

Answer 22

@joyraheb

Answer 23

ok so use delta=b^2 -4ac

Answer 24

What is that? idk what delta is