Question 472862: I have to find the x and y intercepts axis of symmentary and vertex of this equation
y= -x^2/2+3x+8
to find x intercepts told to mutiply by -2, which makes equation
y=x^-6x-16
would I find axis of symmentry and vertex from the new values or from original equation.
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
y= -x^2/2+3x+8
0 = -x^2/2+3x+8 | x-intercepts when y = 0
0 = x^2 - 6x -16 = (x-8)(x+2) x-intercepts (8,0)(-2,0)
the vertex form of a parabola,  where(h,k) is the vertex
y= -x^2/2+3x+8 |complete the square of the original EQ
y= -(1/2)[ x-3)^2 -9] + 8
y= -(1/2)[ x-3)^2 + 25/2 V(3,12.5)
Answer by MathLover1(20849)  (Show Source):
You can put this solution on YOUR website! The  occurs when  , make the equation =  and solve for 
 ......... multiply by 
 .......or
 .........use quadratic formula
solutions:
or
 occurs when  , substitute for in the equation, and find 
so, the  are at points ( ,) and ( ,)
the is at point ( )
Axis of symmetry can be found using the formula: 
since  and  ,
so, the axis of symmetry is 
Vertex is the x/y values for the max or min and occurs at the axis of symmetry; so Substitute  for and find the value:
so, the Vertex is at (, )
now see it on a graph:
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