Question

If the function h is defined by h(x)=
`x^(2)`-3x+5,then h(2x+1)=

Answer 1

Given:

The function is:

`h(x)=x^2-3x+5`

To find:

The value of
`h(2x+1)`.

Solution:

We have,

`h(x)=x^2-3x+5`

Putting
`x=2x+1`, we get

`h(2x+1)=(2x+1)^2-3(2x+1)+5`

`h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5`

`h(2x+1)=4x^2+4x+1-6x-3+5`

On combining like terms, we get

`h(2x+1)=4x^2+(4x-6x)+(1-3+5)`

`h(2x+1)=4x^2-2x+3`

Therefore, the required function is
`h(2x+1)=4x^2-2x+3`.