What is the standard form of #y= (7/5x-4/7)^2+4#?

1 Answer
Oct 2, 2017

#y=49/25x^2 -8/5x + 212/49#

Explanation:

Basically you just expand out the bracket.

Rule for squaring things: the first squared, plus the last squared, plus twice the product of the two. (like if you had #(x+3)^2# it'd be #x^2 + 3^2 + "twice" (3*x) = x^2+6x+9#)

So, #(7/5x-4/7)^2# will be #(7/5x)^2# + #(-4/7)^2# + #2(7/5x*-4/7)#
#=49/25x^2 + 16/49 -8/5x#

Now add the +4:

#=49/25x^2 -8/5x + 4 + 16/49 = 49/25x^2 -8/5x +212/49#