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

Jul 27, 2016

Domain: $\left\{x : x = N\right\}$ Where $N$ is the set of all real numbers.
Range: $\left\{p \left(x\right) : p \ge 9\right\}$

#### 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 $p \left(x\right)$.

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

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

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

Method 2 (Differentiation)
$\frac{d}{\mathrm{dx}} {x}^{2} - 2 x + 10 = 2 x - 2$

Find the turning point: $\frac{d}{\mathrm{dx}} p \left(x\right) = 0$ hence $2 x - 2 = 0$ so $x = 1$.

$p \left(x = 1\right) = {1}^{2} - 2 + 10 = 9$ This is your minimum point.

From Method 1 and 2, your minimum point is 9. So the range is $\left\{p \left(x\right) : p \ge 9\right\}$