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

1 Answer
Feb 16, 2018

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

Explanation:

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!