Determine all x-intercepts, vertical asymptotes, horizontal asymptotes, and removable discontinuities of f (x)=(x^2-x-6)/(x^3-4x).
recall that vertical asymptotes occur when the denominator is zero and the numerator is not.
Removable discontinuities are where the fraction is 0/0
Horizontal asymptotes are what happens when x gets huge.
f(x) =
(x-3)(x+2)
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x^2 (x-4)
get it now? If not, where do things get creepy?
To determine the x-intercepts, we need to find the values of x for which f(x) is equal to zero. So let's start by solving the equation:
(x^2 - x - 6) / (x^3 - 4x) = 0
To find the x-intercepts, we need to find the values of x for which the numerator is equal to zero since a fraction is equal to zero if and only if its numerator is equal to zero. So let's solve the numerator equation:
x^2 - x - 6 = 0
This can be factored as:
(x-3)(x+2) = 0
Setting each factor equal to zero and solving for x, we get:
x-3=0, which gives x=3
x+2=0, which gives x=-2
Therefore, the x-intercepts of f(x) are at x=3 and x=-2.
Next, let's determine the vertical asymptotes. A vertical asymptote occurs when the denominator of the fraction is equal to zero, excluding any values where the numerator is also equal to zero (as those are x-intercepts). So let's solve the denominator equation:
x^3 - 4x = 0
Factoring out an x, we have:
x(x^2 - 4) = 0
Setting each factor equal to zero, we get:
x=0 (which we already know is an x-intercept)
x^2 - 4 = 0
Solving x^2 - 4 = 0, we can factor this as:
(x-2)(x+2) = 0
Setting each factor equal to zero and solving for x, we get:
x-2=0, which gives x=2
x+2=0, which gives x=-2
Therefore, the vertical asymptotes of f(x) are at x=2 and x=-2. Note that x=-2 is excluded as it is already an x-intercept.
Now let's look for any removable discontinuities. These occur when some factors in the numerator and denominator cancel each other out, leaving a hole in the graph. In this case, there are no factors that cancel each other out, so there are no removable discontinuities.
Lastly, let's determine the horizontal asymptotes. To find horizontal asymptotes, we need to look at the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is equal to the degree of the denominator, we need to compare the leading coefficients. If the leading coefficients are equal, the horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator). If the leading coefficients are not equal, there is no horizontal asymptote.
In this case, the degree of the numerator is 2, and the degree of the denominator is 3. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.
To summarize:
- x-intercepts: x=3, x=-2
- Vertical asymptotes: x=2 (x=-2 is excluded as it is an x-intercept)
- Removable discontinuities: None
- Horizontal asymptotes: y=0