Identify the x-int and the vertex of the quadratic function

Question

f(x)=2x^2-x+1

 

what I did:

2x^2/2 - x/2 = -1/2

x^2 - x/2 = -1/2

x^2 - x/2 +1/16 = -1/2 +1/16

(x-1/4)^2= -7/16

Vertex: (1/4, 7/16)

But when I graph it the vertex is incorrect

what is the equation in complete the square?

Answer 1

What you did is set f(x) equal to zero and added a number on both sides of the equation to get a complete square on one side of the equation.  This will get you the x-intercepts, but not the y-intercepts and the vertex.

 

 

Write the given function in vertex form:

 

f(x) = a(x - h)² + k

 

where:

a is the coefficient of the x² term

The coordinate of the vertex is (h, k)

 

 

f(x) = 2(x² - (1/2)x + 1/16) + (7/8)

 

f(x) = 2(x - 1/4)(x - 1/4) + (7/8)

 

f(x) = 2(x - 1/4)² + (7/8)

 

 

The vertex is (1/4, 7/8).

 

 

 

 

Answer 2

It looks to me that you did it right, and when I graph the equation I see the vertex where you say it should be. 

 

Try graphing again -- maybe your graph wasn't right.