What is the range of the function #y=2x^2 +32x - 4#?

1 Answer
Jan 17, 2018

Answer:

#y>=-132#

Explanation:

This a quadratic function with positive leading coefficient so it has a minimum value at its vertex. The vertex of #y=ax^2+bx+c# is #(h,k)# where #h=-b/(2a)# and #k# is found by substitution.

For the given function #h=-32/(2*2)=-32/4=-8#.

Find #k# by substitution:

#k= 2(-8)^2+32(-8)-4#

#k= 128-256-4#
#k=-132#

Since the minimum is #(-8,-132)# and there is no maximum value, the range of the function is #y>=-132#.