How do you find the equation of a parabola when given a vertex (3, -3) point (0, 6)?

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Answer 1 · 2015-07-27T17:28:06

I found: #y=x^2-6x+6#

You have
1] the general equation of your parabola:
#y=ax^2+bx+c " " (1)#
2] two points of your parabola:

#x=3, y=-3# which is also the vertex. At the vertex #x_v=3=-b/(2a)#;

#x=0, y=6#

we can try use these information into equation (1):

#-3=9a+3b+c#

#6=0a+0b+c# so that #color(red)(c=6)#

#3=-b/(2a)# and #b=-6a#

So we can get, from the first (with #c=6# and #b=-6a#):
#-3=9a-18a+6# and #color(red)(a=1)#
and so: #b=-6a=color(red)(-6)#

The equation of the parabola is:
#y=x^2-6x+6#
Graphically:
graph{x^2-6x+6 [-16.02, 16.02, -8.01, 8.01]}