What is the vertex form of #y= 2x^2 + x - 1 #?

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Answer 1 · 2018-02-16T02:58:07

#y=2(x+1/4)^2-1 1/8# with the vertex as #(-1/4,-1 1/8)#

Vertex form is given by

#y=a(x-h)^2+k# with #(h,k)# being the vertex.

To get to vertex form, complete the square.

#y=2(x^2+1/2xcolor(red)(+(1/4^2)-(1/4^2)))-1#

#y=2(x+1/4)^2-1/2-1#

#y=2(x+1/4)^2-1 1/8# and the vertex is #(-1/4,-1 1/8)#

Note: If you want to just find the vertex using #ax^2+bx+c#:

#h=-b/(2a)#

#k=c-b^2/(4a)#

with #(h,k)# being the vertex:

#h=(-1)/(2*2)#

#h=-1/4#

#k=-1+(1^2)/(4*2*-1)#

#k=-1 1/8#

#(h,k)# is #(-1/4,-1 1/8)#

If you were to find vertex form using the method previously described, then you would need a point on the graph of #2x^2+x-1#.

Let's say that they told you #(-1,0)# was a point on the graph.

Using the vertex as found out with the formulas above, plug into vertex form with what you have:

#y=a(x+1/4)^2-1 1/8#

To find out #a#, plug the point #(-1,0)# into what you have:

#0=a(-1+1/4)^2-1 1/8#

#0=9/16a-1 1/8#

#1 1/8=9/16a#

#18/16=9/16a#

#a=2# giving you

#y=2(x+1/4)^2-1 1/8#

Here is a graph for reference: graph{2x^2+x-1 [-10, 10, -2, 5]}

Hope this helps!