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November 24, 2014

November 24, 2014

Posted by **David** on Saturday, March 19, 2011 at 7:52pm.

y=3x^2 + 12x + 9

12/6 = -2 = y

3(-2)^2 + 12(-2) + 9 = -3

x=0 y=9 (0,9)

Would I just plug in 1 and 2 for x and then reflect it while i'm graphing?

So, if I do that, should it come out...

x=1

-----

y=3(1)^2 + 12(1) + 9

y=3+12+9

y=24 (1,24)

x=2

------

y=3(2)^2 + 12(2) + 9

y=12+24+9

y=45 (2,45)

Would this be correct?

- Algebra -
**bobpursley**, Saturday, March 19, 2011 at 8:14pmyou are correct on the points 1,24 adn 2,45

The first part, I do not understand what you were doing.

- Algebra -
**Damon**, Saturday, March 19, 2011 at 8:17pmy=3x^2 + 12x + 9

you are correct that when x = 0, y = 9 so (0,9) is on it

However I do not know what else you are doing.

I generally complete the square to find the axis of symmetry

3x^2 + 12x + 9 = y

get one by x^2 so divide by 3

x^2 + 4 x + 3 = y/3

subtract 3 from both sides

x^2+4x = y/3 - 9/3 =

add (half of 4)^2 to both sides

x^2 + 4 x + 4 = y/3 -9/3 + 12/3

(x+2)^2 = (1/3)(y+3)

vertex at (-2, -3)

vertical axis of symmetry at x = -2

- Algebra -
**David**, Saturday, March 19, 2011 at 8:32pmWell, I'm trying to find two different points by changing the x value since I need to graph these points and then reflect it. I hope this helps you a little.

- Algebra -
**Damon**, Saturday, March 19, 2011 at 8:53pmok, well reflect it about the line x = -2

and the bottom is at (-2,-3)

Your (1,24) and your (2,45) are fine

so

first point is the bottom (-2,-3)

next is your (1,24)

the reflection of that about x = -2 is

3 to the left of -2 which is (-5,24)

next do your (2,45)

that is four to the right of the line of symmetry so go four to the left of -2 to -6

so (-6,45)

- Algebra -
**Damon**, Saturday, March 19, 2011 at 8:57pmthis might help:

http://www.purplemath.com/modules/sqrvertx.htm

- Algebra -
**Damon**, Saturday, March 19, 2011 at 8:59pmand this:

http://uncw.edu/courses/mat111hb/pandr/quadratic/quadratic.html

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