How do you find the minimum and maximum value of #y=2(x-3)(x-5)#?

1 Answer
May 16, 2017

Maximum value is #(oo)# , minimum value is #-2 #

Explanation:

#y=2(x-3)(x-5) = 2x^2 -16x +30 :. a = 2, b= -16 , c= 30; [y=ax^2+bx+c]#

Discriminant #D= b^2-4ac = 256-240=16 #

Here roots are real #x=3 and x=5 and a>0#

When roots are real and #a>0# , Maximum value is #(oo)# and
minimum value is #-D/(4a) = -16/(4*2) = -2 #. This is also confirmed from graph. graph{2x^2-16x+30 [-10, 10, -5, 5]} [Ans]