B Bobbie New member Joined Sep 8, 2006 Messages 2 Sep 14, 2006 #1 suppose that f(x)=2x^2-3x. simplify f(b)-f(a)/b-a I got f(2b^2-3b)-f(2a^2-3a)/b-a = 2b^2-2a^2-3b+3a/b-a = 2(b-a)(b-a)+(a-b)/ b-a = 2(b-a)+3(a-b) Help [/u]
suppose that f(x)=2x^2-3x. simplify f(b)-f(a)/b-a I got f(2b^2-3b)-f(2a^2-3a)/b-a = 2b^2-2a^2-3b+3a/b-a = 2(b-a)(b-a)+(a-b)/ b-a = 2(b-a)+3(a-b) Help [/u]
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Sep 14, 2006 #2 \(\displaystyle \L\\\frac{(2b^{2}-3b)-(2a^{2}-3a)}{b-a}=\frac{2b^{2}-3b-2a^{2}+3a}{b-a}=\frac{(b-a)(2a+2b-3)}{b-a}\)
\(\displaystyle \L\\\frac{(2b^{2}-3b)-(2a^{2}-3a)}{b-a}=\frac{2b^{2}-3b-2a^{2}+3a}{b-a}=\frac{(b-a)(2a+2b-3)}{b-a}\)
