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

1 Answer
Jul 27, 2016

Answer:

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}#