Vertex form is given by
y=a(x-h)^2+ky=a(x−h)2+k with (h,k)(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)))-1y=2(x2+12x+(142)−(142))−1
y=2(x+1/4)^2-1/2-1y=2(x+14)2−12−1
y=2(x+1/4)^2-1 1/8y=2(x+14)2−118 and the vertex is (-1/4,-1 1/8)(−14,−118)
Note: If you want to just find the vertex using ax^2+bx+cax2+bx+c:
h=-b/(2a)h=−b2a
k=c-b^2/(4a)k=c−b24a
with (h,k)(h,k) being the vertex:
h=(-1)/(2*2)h=−12⋅2
h=-1/4h=−14
k=-1+(1^2)/(4*2*-1)k=−1+124⋅2⋅−1
k=-1 1/8k=−118
(h,k)(h,k) is (-1/4,-1 1/8)(−14,−118)
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-12x2+x−1.
Let's say that they told you (-1,0)(−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 1/8y=a(x+14)2−118
To find out aa, plug the point (-1,0)(−1,0) into what you have:
0=a(-1+1/4)^2-1 1/80=a(−1+14)2−118
0=9/16a-1 1/80=916a−118
1 1/8=9/16a118=916a
18/16=9/16a1816=916a
a=2a=2 giving you
y=2(x+1/4)^2-1 1/8y=2(x+14)2−118
Here is a graph for reference: graph{2x^2+x-1 [-10, 10, -2, 5]}
Hope this helps!
