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First factor everything
(x - 2)(x^2 +2x +4)
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(x - 3)(x - 2)
Normally, if you have to find these values, there is no common factor in the top and bottom. You should check with your teacher whether he/she wants you to cancel the x - 2 before you do any of the rest of this.
VERTICAL ASYMPTOTES - what x values would make the bottom zero?
Solve x - 3 = 0 and x - 2 = 0 to find the equations of your vertical asymptotes: x = 3 and x = 2
HORIZONTAL ASYMPTOTES - I assume you are not in calculus . . . so, the only way is to compare the degrees of the top and bottom, then, if the bottom is higher than the top, the H.A. is y = 0, if the degrees are equal, the H.A. is the ratio of the leading coefficients. If the degree of the top is bigger than the bottom, you have no H.A. Instead, you have an angled or oblique asymptote.
Y-INTERCEPT: Set x = 0 and then solve. I assume your question had a 'y =' in front of it. (or f(x))
y = (0-8)/(0-0+6) = -8/6 = -4/3 so the y-int is the point (0, -4/3)
X-INTERCEPT: Set y = 0 and solve. To solve, realize the only way a fraction can equal zero is if the numerator equals zero. So you just need to figure out what makes the top zero. Note that the trinomial factor of a difference of cubes never can equal zero, so for this question you only need to work with the x - 2.
x - 2 = 0
x = 2
So the x intercept is (2, 0).
The DOMAIN of a normal polynomial function is x E R. This means x can be anything real. For a fraction polynomial like this one, the x's in the bottom create 'restrictions' on this domain at any value that would make the bottom zero. So, at your vertical asymptotes, your domain is restricted. For this function, then, your domain is
x is not equal to 2, x is not equal to 3, x E R (Instead of 'not equal to', please use an equal sign with a slash through it.)
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