If you entered it as you typed it here, then you need some parentheses
(9x^3+9x^2−9x−9)/ (x^2−x−2)
= 9(x^3+x^2-x-1) / (x^2-x-2)
= 9(x-1)(x+1)^2 / (x-2)(x+1)
= 9(x^2-1)/(x-2)
you know that y=0 at x = -1 and 1
for x in the interval (-1,1), y>0
for x>2, y>0
y<0 everywhere else
Sketch the graph of the given rational function. Use the graph to complete the following. If they do not exist or if any blanks are empty, enter DNE in all capital letters.
f(x)=9x^3+9x^2−9x−9/ x^2−x−2
Interval(s) on which f(x)>0:
Interval(s) on which f(x)<0:
I tried using a graphing calculator....I know that everything greater than zero is above the x axis and visa versa for the f(x)<0
BUT
I don't know if I'm entering the format wrong or what
thanks
2 answers
9(x^3+x^2-x-1) /[ (x-2)(x+1)]
=9(x+1)(x^2-1) /[ (x-2)(x+1)]
=9(x^2-1) / (x-2)
= 9 (x-1)(x+1) / (x-2) explodes if x = +2, 0 if x = +1 or -1
if x > +2, all terms are +
if 1 < x < 2 top + and bottom - so negative
if -1 < x <+1 top - and bottom - so positive
if x < -1, top - and bottom - so positive
CHECK THAT = I did it fast
=9(x+1)(x^2-1) /[ (x-2)(x+1)]
=9(x^2-1) / (x-2)
= 9 (x-1)(x+1) / (x-2) explodes if x = +2, 0 if x = +1 or -1
if x > +2, all terms are +
if 1 < x < 2 top + and bottom - so negative
if -1 < x <+1 top - and bottom - so positive
if x < -1, top - and bottom - so positive
CHECK THAT = I did it fast