What is the maximum value of #y = - 9x^2 +126x-431# ?

1 Answer
Jan 15, 2016

#y_max = 893# at #x_max=7#

Explanation:

The maximum values of a quadratic #y=ax^2 + bx +c# is found at #x=-b/(2a) #

In this example #a=-9# and #b=126#
#:.x_max = (-126)/-18 = 7#

Plugging this back into the original equation gives
#y_max = -9*7^2 +126*7 -431 = 441 + 883 -431#
#y_max = 893#