Posted by David on Saturday, March 19, 2011 at 7:52pm.
Hello. I am trying to solve quadratic functions but I'm having difficulty understanding. Here is my work so far...
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 8:14pm
you are correct on the points 1,24 and 2,45
The first part, I do not understand what you were doing.

Algebra  Damon, Saturday, March 19, 2011 at 8:17pm
y=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 8:32pm
Well, 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 8:53pm
ok, 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 8:57pm
this might help:
http://www.purplemath.com/modules/sqrvertx.htm

Algebra  Damon, Saturday, March 19, 2011 at 8:59pm
and this:
http://uncw.edu/courses/mat111hb/pandr/quadratic/quadratic.html
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