How do you find the local max and min for #f(x)= x^2 + 8x -12#? Calculus Graphing with the First Derivative Classifying Critical Points and Extreme Values for a Function 1 Answer Eddie Jun 22, 2016 #y_{min} = -28# at #x = -4# Explanation: calculus is OTT for this So #y = x^2 + 8x - 12 = (x+4)^2 - 16 - 12# #= (x+4)^2 - 28# #(x+4)^2 \ge 0# so #y_{min} = -28# at #x = -4# Answer link Related questions How do you find and classify the critical points of #f(x)=x^3#? How do you find the critical points of a rational function? How do you know how many critical points a function has? How many critical points can a cubic function have? How many critical points can a function have? How many critical points can a quadratic polynomial function have? What is the first step to finding the critical points of a function? How do you find the absolute extreme values of a function on an interval? How do you find the extreme values of the function and where they occur? What is the extreme value of a quadratic function? See all questions in Classifying Critical Points and Extreme Values for a Function Impact of this question 1750 views around the world You can reuse this answer Creative Commons License