The given function is
f(x) = x³ - 11x² + 36x - 36
When x = 0, f = -36
The y-intercept is -36
Possible zeros are +/-1, +/-2, +/-3, +/-4, +/-6, +/-9
From the Remainder Theorem, x=a is a zerp if f(a) = 0.
Try x = 1:
f(1) = 1 - 11 + 36 - 36 = -10 (not a zero)
Try x = 2:
f(2) = 8 - 44 + 72 - 36 (this is a zero).
Use synthetic division.
2| 1 -11 36 -36
2 -18 36
----------------------
1 -9 18 0
Therefore
f(x) = (x - 2)(x² - 9x + 18)
= (x - 2)(x - 3)(x - 6)
The x-intercepts are x = 2, 3, 6
As x→ -∞, f→ -8
As x→ +∞, f→ +∞
To find how f(x) behaves between the zeros, test function values between zeros.
x f(x) Comments
---- --------- ---------------
-1 -84 negative
0 -36 negative
2 0 zero
2.5 0.875 positive
3 0 zero
4 -4 negative
6 0 zero
8 60 positive
Summary:
y-intercept = -36
x-intercepts: 2.3.6
f→-∞ as x→-∞
f→+∞ as x→+∞
We now know enough about the shape of the curve to sketch it, as shown below.
