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?
you are correct on the points 1,24 and 2,45
The first part, I do not understand what you were doing.
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
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.
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)
this might help:
http://www.purplemath.com/modules/sqrvertx.htm
and this:
http://uncw.edu/courses/mat111hb/pandr/quadratic/quadratic.html
To solve quadratic functions, you need to follow a specific process. Let's go through it step by step using your example:
The quadratic function you provided is y = 3x^2 + 12x + 9.
Step 1: Find the vertex of the parabola.
To find the vertex, you can use the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c. In your case, a = 3, b = 12, and c = 9.
x = -12/(2*3) = -12/6 = -2
To find the y-coordinate of the vertex, substitute the x-value (-2) back into the original equation:
y = 3(-2)^2 + 12(-2) + 9
y = 12 - 24 + 9
y = -3
So, the vertex of the parabola is (-2, -3).
Step 2: Find the y-intercept.
To find the y-intercept, substitute x = 0 into the equation:
y = 3(0)^2 + 12(0) + 9
y = 0 + 0 + 9
y = 9
So, the y-intercept is (0, 9).
Step 3: Find additional points.
To graph the parabola, you can either plug in specific x-values and calculate the corresponding y-values or use symmetry around the vertex.
If you choose to plug in specific x-values, you can calculate additional points like you did:
For x = 1:
y = 3(1)^2 + 12(1) + 9
y = 3 + 12 + 9
y = 24
So, the point is (1, 24).
For x = 2:
y = 3(2)^2 + 12(2) + 9
y = 12 + 24 + 9
y = 45
So, the point is (2, 45).
If you choose symmetry, you can reflect the point (0, 9) over the vertex (-2, -3) to find another point on the other side of the parabola. This would be (4, 9).
Both methods are correct, and both points (1, 24) and (2, 45) you calculated are correct.
By plotting these points on a graph, you can draw the parabola for the quadratic function y = 3x^2 + 12x + 9.
I hope this explanation helps you understand how to solve quadratic functions and graph the corresponding parabola. If you have any more questions, feel free to ask!