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
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
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
To sketch the graph of the given rational function, you can follow these steps:
1. Factor the numerator and denominator to find any x-intercepts and vertical asymptotes, if they exist. In this case, the numerator is 9x^3 + 9x^2 - 9x - 9, which can be factored as 9(x^2 - 1)(x + 1), and the denominator is x^2 - x - 2, which factors as (x - 2)(x + 1). So the function has a vertical asymptote at x = 2 and x = -1, and x-intercepts at x = -1 and x = 1.
2. Determine the behavior of the function as x approaches both positive and negative infinity. You can do this by comparing the degrees of the numerator and denominator. In this case, the degree of the numerator (3) is greater than the degree of the denominator (2), so as x approaches positive or negative infinity, f(x) will also approach positive or negative infinity, respectively.
3. Find any horizontal or slant asymptotes, if they exist. In this case, since the degree of the numerator is greater than the degree of the denominator by 1, the rational function does not have a horizontal asymptote. Therefore, there is no horizontal asymptote to consider.
4. Plot the x-intercepts, vertical asymptotes, behavior at infinity, and any additional points if necessary to sketch the graph of the rational function.
Given that f(x) > 0 represents the intervals where the graph is above the x-axis and f(x) < 0 represents the intervals where the graph is below the x-axis, we can use the graph to find these intervals:
From the factorization in step 1, we found that the x-intercepts are x = -1 and x = 1. Therefore, these are the potential intervals in which f(x) can change sign.
By analyzing the graph, we can see that f(x) is positive (above the x-axis) between x = -1 and x = 1. Hence, the interval on which f(x) > 0 is (-1, 1).
Since the graph has a vertical asymptote at x = -1 and x = 2, the graph will be negative (below the x-axis) in the intervals (-∞, -1) and (1, 2).
Therefore, the interval(s) on which f(x) < 0 is (-∞, -1) and (1, 2).
It's worth noting that while a graphing calculator can help visualize the graph, it's always useful to understand the steps involved in sketching the graph and determining intervals algebraically as well.