SOLUTION: I was absent and my friends notes are off so I'm really confused. Your help would be greatly appreciated. Here are my instructions : Write the equation of the axis of symmetry, a

Question 418953: I was absent and my friends notes are off so I'm really confused. Your help would be greatly appreciated. Here are my instructions : Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each function. Idetify the vertex as a maximum or minimum. Then graph the function. The first ones are REALLY HARD. However, I'll give an easy one so I can see it all done so I can figure out the hard ones. The problem is :
y = 3x^2 - 6x + 4 . We are graphing quadratic functions. Thanks so much!
Found 2 solutions by shree840, stanbon:
Answer by shree840(260) (Show Source): You can put this solution on YOUR website!
I used ti 84 to graph. The graph does not touch x axis so these are imaginary roots. Vertex at x=1 y=1. Line of symmetry equation is x=1. Since the graph is upward, it has a minima at x=1, y=1.
You may be being asked to graph by hand.
set your table for x and y
put some values for x and find correesponding value for y
graph the coordinates.
for example x=0,1,2,3,-1,-2,-3
hope you get this.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each function. Idetify the vertex as a maximum or minimum. Then graph the function. The first ones are REALLY HARD. However, I'll give an easy one so I can see it all done so I can figure out the hard ones. The problem is :
y = 3x^2 - 6x + 4
----
Since 3 is positive the parabola opens up.
It therefore has a minimum point.
That point occurs when x = -b/2a = 6/(2*3) = 1
---
The corresponding y-value is f(1) = 3-6+4 = 1
-
So the vertex is (1,1)
----
The axis of symmetry is a vertical line thru the vertex.
Its equation is x = 1.
---
==========================================
Cheers,
Stan H.
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