Posted by leticia on Tuesday, March 1, 2011 at 12:47pm.
We have a fraction in which a cubic Eq is the numerator and a Quadratic Eq is the denominator. Let's factor both.
Find a value of X that gives 0 out when
plugged into the cubic Eq. It was found
by trial and error that -2 will give
an output of 0 and,therefore, satisfies
the Eq.
X = -2,
X + 2 = 0. Using long-hand division,
divide the cubic Eq by x + 2 and get:
X^2 - 1.
(X + 2)(X^2 - 1) = 0,
(X + 2)(X + 1)(X - 1) = 0 = numerator.
Factor the denominator and get:
(X - 2)(X + 3) = 0. Therefore,
F(x)=(X + 2)(X + 1)(X -1)/(X -2)(X + 3)
The domain = all real values of x
that does not make the denominator = 0.
2 and -3 will make the denominator = 0
and are outside the domain.
1. So the domain = all real values of X
except 2 and -3.
3. a. Not in lowest term as given.
b. Proper fraction.
4. X-Ints = -2, -1 , 1 = roots = solu.
When X = 0, F(0) = -2/-6 = 1/3 = Y-int.
5. Zeroes = X-Ints = -2,-1, and 1.
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