Question

Determine the domain and the range of the function.C. Determine where the function is increasing and where it is decreasing.

Answer 1

Given:

`f(x)=2x^2-x+1`
`a=2\text{ ; b= -1 ; c=1}`

Graph opes upwards.

Let the vertex be (h,k)

`h=-(b)/(2a)`
`h=-((-1))/(2(2))`
`h=(1)/(4)`
`k=f(h)`
`k=2((1)/(4))^2-(1)/(4)+1`
`k=2((1)/(16))-(1)/(4)+1`
`k=(1)/(8)-(1)/(4)+1`
`k=(1-2+8)/(8)`
`k=(7)/(8)`
`\text{Vertex}=((1)/(4),(7)/(8))`

Axis of symmetry is

`x=(1)/(4)`

y- intercept

x=0,

y=1

There is no x intercept .

Domain:

`(-\infty,\infty)`

Range:

`\lbrack(1)/(4),\infty)`

The function is increasing:

`((1)/(4),\infty)`

The function is decreasing:

`(-\infty,(1)/(4))`