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...
y=3(1)^2 + 12(1) + 9
y=3(2)^2 + 12(2) + 9
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
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
- Algebra - Damon, Saturday, March 19, 2011 at 8:57pm
this might help:
- Algebra - Damon, Saturday, March 19, 2011 at 8:59pm
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