Hi,  

How do you graph 

Y = x^2 -6x +8   || Putting into Vertex Form by completing the Square

y =(x-3)^2 - 9 + 8

y = (x-3)^2 - 1    and when (x-3)^2 - 1 = 0 ,,  x = 3 ± 1

the vertex form of a Parabola opening up(a>0) or down(a<0),  

where(h,k) is the vertex  and  x = h  is the Line of Symmetry

Parabola opening Upward, V(3,-1) and x - 3 the line of symmetry

x-intercepts (2,0) and (4,0)



 

Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
 



A parabola is of the general form , and you have that.



If  is  the parabola  (looks like a smiley face), if  is  the parabola  (looks like a sad face).



It is  to note that for your question, you have the opposite , instead of , you goes up. 



In case of  will be on , and when this occurs, a positive  value means it will go to the , so it will look like a letter .





first, you can do is to find the vertex using 



 ...this is  value...plug it in  and find  value



















so, the vertex is at the point (,)





In this example you can factor to find the , the possible values for  in our general form, when . 



So,

 

........substitute  with  



...group



...factor out  from first group and  from second group



...set 







if ..->...



if ..->...





So the parabola passes through the  on the graph at  and . As parabola are symmetrical, the  of  is halfway between these, ie . 



So sub this value into the original equation to find the lowest point of the parabola.



...now set  and solve for  to find 







...so, there is  at the point (,)



you can also take few other positive and negative values for  , calculate  and plot that points too, and you are ready to draw a graph

 





 





 

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