What is the maximum value that the graph of #y= x^2 -4x#? Calculus Graphing with the First Derivative Identifying Stationary Points (Critical Points) for a Function 1 Answer Jim H Oct 6, 2016 There is no maximum. Explanation: As #x# increases without bound, #x^2-4x# also increases without bound. (Another way of saying this is: #lim_(xrarroo)(x^2-4x) = oo#.) That means there is no upper bound and no maximum. Answer link Related questions How do you find the stationary points of a curve? How do you find the stationary points of a function? How many stationary points can a cubic function have? How do you find the stationary points of the function #y=x^2+6x+1#? How do you find the stationary points of the function #y=cos(x)#? How do I find all the critical points of #f(x)=(x-1)^2#? Let #h(x) = e^(-x) + kx#, where #k# is any constant. For what value(s) of #k# does #h# have... How do you find the critical points for #f(x)=8x^3+2x^2-5x+3#? How do you find values of k for which there are no critical points if #h(x)=e^(-x)+kx# where k... How do you determine critical points for any polynomial? See all questions in Identifying Stationary Points (Critical Points) for a Function Impact of this question 1388 views around the world You can reuse this answer Creative Commons License