How do you find the domain and range of p(x)=x2−2x+10?

1 Answer
Alexander L.
Jul 27, 2016

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 x2 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 x2−2x+10=(x+a)2+c.
Expand the right hand side of the equation and you will get:
x2−2x+10=x2+2ax+a2+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)
ddxx2−2x+10=2x−2

Find the turning point: ddxp(x)=0 hence 2x−2=0 so x=1.

p(x=1)=12−2+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}

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