What is the domain and range of `f(x)= x^2 - 6x + 8`?

Domain:

`f(x)` is a quadratic equation and any values of `x` will give a real value of `f(x)`.

The function does not converge to a certain value ie: `f(x)=0` when `x->oo`

Your domain is `{x: -oo<=x<=oo}`.

Range:

Method 1-
Use completing the square method:
`x^2-6x+8=(x-3)^2-1`
Hence you minimum point is `(3,-1)`. It is a minimum point because the graph is a "u" shape (coefficient of `x^2` is positive).

Method 2-
Differentiate :
`(df(x))/(dx)=2x-6`.

Let`(df(x))/(dx)=0`

Therefore, `x=3` and `f(3)=-1`
Minimum point is `(3,-1)`.
It is a minimum point because the graph is a "u" shape (coefficient of `x^2` is positive).

Your range takes values between `-1 and oo`

It is a polynomial function, its domain is all real numbers. In interval notation this can be expressed as `(-oo, +oo)`
For finding its range, we can solve the equation y= `x^2-6x+8` for x first as follows:

`y= (x-3)^2 -1`,
`(x-3)^2 = y+1`
x-3= `+-sqrt(y+1)`
x= 3`+-sqrt(y+1)`. It is obvious from this that y`>=-1`

Hence range is `y>=-1`. In interval notation this can be expressed as`[-1, +oo)`