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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