How do you determine if #f(x) = (x - 2)^2 + 1# is an even or odd function? Precalculus Functions Defined and Notation Introduction to Twelve Basic Functions 1 Answer Shwetank Mauria Aug 31, 2016 #f(x)# is neither odd nor even. Explanation: If a function #f(x)# is even than #f(-x)=f(x)# and if it is odd than #f(-x)=-f(x)#. As #f(x)=(x-2)^2+1#, #f(-x)=(-x-2)^2+1# = #(x+2)^2+1# and hence #f(-x)# is neither equal to #f(x)# nor equal to #-f(x)#. Hence, #f(x)# is neither odd nor even. Answer link Related questions What are the twelve basic functions? What is the greatest integer function? What is the absolute value function? What is the graph of the greatest integer function? What is the graph of the absolute value function? What is the inverse function? What is the graph of the inverse function? Which of the twelve basic functions are bounded above? Which of the twelve basic functions are their own inverses? How do you use transformations of #f(x)=x^3# to graph the function #h(x)= 1/5 (x+1)^3+2#? See all questions in Introduction to Twelve Basic Functions Impact of this question 1209 views around the world You can reuse this answer Creative Commons License