Question

How do you find the domain and range of `p(x)=x^2-2x+10`?

Answer 1

Domain: `{x: x=N}` Where `N` is the set of all real numbers.
Range: `{p(x): p>=9}`

Explanation:

We know that the polynomial is a "U" shape because the coefficient of the `x^2` term is positive hence it has a minimum value.

Domain
This is a polynomial hence all real values of `x` are valid for `p(x)`.

Range
Method 1 (Completing the Square):
Let `x^2-2x+10=(x+a)^2+c`.
Expand the right hand side of the equation and you will get:
`x^2-2x+10=x^2+2ax+a^2+c`

Hence `a=-1` and `c=9`

Since `p(x)=(x-1)^2+9`, it has minimum value (which equals to 9) if `(x-1)^2=0`. Therefore the range is anything from 9 and beyond.

Method 2 (Differentiation)
`d/(dx)x^2-2x+10=2x-2`

Find the turning point: `d/(dx)p(x)=0` hence `2x-2=0` so `x=1`.

`p(x=1)=1^2-2+10=9` This is your minimum point.

From Method 1 and 2, your minimum point is 9. So the range is `{p(x): p>=9}`