How do you find a quadratic function #f(x) = ax² + bx + c# given range# {y:y ≤ 9}#; #x#-intercepts -2 and 4?

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Answer 1 · 2016-09-30T12:24:55

#a = -9/8, b = 9/4, c = 9#

Regarding the intercepts

#f(-2) = (-2)^2a+(-2)b+c = 0#
and
#f(4)=4^2a+4b + c = 0#

The range stablishes that for #-2 le x le 4# then
#y ge 0# having it's maximum at #x = (-2+4)/2 = 1# so the last equation is

#f(2) = 2^2a+2b+c = 9#

Solving now

#{( (-2)^2a+(-2)b+c = 0),(4^2a+4b + c = 0),(2^2a+2b+c = 9):}#

we obtain

#a = -9/8, b = 9/4, c = 9#