How do you evaluate the function #p(x)=1/2x^3 + 2/3x^2 - 1/4x + 1/3# for #p(-2)#? Algebra Expressions, Equations, and Functions Function Notation 1 Answer Lawrence G. Mar 24, 2018 #p(-2)=1/2(-2)^3+2/3(-2)^2-1/4(-2)+1/3# #=1/2(-8)+2/3(4)+2/4+1/3# #=-8/2+8/3+2/4+1/3# #=-4+(8/3+1/3)+1/2# #=-4+9/3+1/2# #=-4+3+1/2# #=-1+1/2# #=-2/2+1/2# #=(-2+1)/2# #=-1/2# Explanation: Just put the #x=-2# on the polynomial equation and you will simplify and you will get #p(-2)=-1/2# Answer link Related questions How do you find the output of the function #y=3x-8# if the input is -2? What does #f(x)=y# mean? How do you write the total cost of oranges in function notation, if each orange cost $3? How do you rewrite #s=2t+6# in function notation? How do you find the value of #f(-9)# for #f(x)=x^2+2#? What does a dependent and independent variable mean? What is the difference between an equation written in function notation and one that is not? How do you evaluate function notation? What is Function Notation? How do you write equations in function notation? See all questions in Function Notation Impact of this question 4713 views around the world You can reuse this answer Creative Commons License