How do you find the max or minimum of #f(x)=3/4x^2-5x-2#?

1 Answer
Apr 26, 2017

We have a minimum of #-10 1/3# at #x=10/3#.

Explanation:

To find

first that whether #f(x)=3/4x^2-5x-2# has a maximum or minimum, one has to first look at the coefficient of #x^2#. Here it is positive, so we have a minimum . Note that for this function has to be in the form #f(x)=ax^2+bx+c#.

Secondly to find the minimum value (or maximum in case coefficient of #x^2# is negative), we should convert it into vertex form i.e. #f(x)=a(x-h)^2+k#, where we have a maximum of #k# at #x=h#.

#f(x)=3/4x^2-5x-2#

= #3/4(x^2-20/3x)-2#

= #3/4(x^2-2xx10/3x+(10/3)^2)-(10/3)^2xx3/4-2#

= #3/4(x-10/3)^2-100/12-2#

= #3/4(x-10/3)^2-31/3#

Hence, we have a minimum of #-31/3=-10 1/3# at #x=10/3#.

graph{3/4x^2-5x-2 [-17.41, 22.59, -11.92, 8.08]}