Answer:
Eccentricity = 5/6
Type of conic section; Ellipse
Directrix; x = -11/5
Step-by-step explanation:
The first step would be to write the polar equation of the conic section in standard form by multiplying the numerator and denominator by 1/6;
The polar equation of the conic section is now in standard form;
The eccentricity is given by the coefficient of cos theta in which case this would be the value 5/6. Therefore, the eccentricity of this conic section is 5/6.
The eccentricity is clearly between 0 and 1, implying that the conic section is an Ellipse.
Since the conic section is in standard form, the numerator is the product of eccentricity and the value of the directrix, that is;
e*d = 11/6
5/6*d = 11/6
d = 11/5
Since the denominator has a minus sign then the ellipse opens towards the right and thus the equation of its directrix is;
x = -11/5
F(x) = 2÷(x² - 2x - 3)
1)
Domain:
The domain is all the values for x that will produce a real number for y
Factor the denominator to find where y is not defined:
f(x) = 2÷(x² - 2x - 3)
f(x) = 2 ÷ (x-3)(x+1)
The domain is all real numbers except x=3 and x=-1
Range:
The range for y is all the values that y can take, given the domain.
The range is all real numbers, because y approaches both positive and negative infinite at different points on the graph.
The y-intercept is where x=0
y= 2 ÷ (0² - 2(0) -3)
y= 2 ÷ -3 = -2/3
The x-intercept are the points at which y=0.
Let's use the factored form again:
f(x) = 2 ÷ (x-3)(x+1)
This function has no x-intercepts. All values of X either produce a real number, or are undefined in the case of x=3 and x=-1
Horizontal Asymptotes
As X approaches inifinite, how does y behave?
f(x) = 2÷(x² - 2x - 3)
As x approaches both positive and negative infinite, the dominate term in the denominator, x², is vastly greater than 2, and thus y approaches zero.
The horizontal asymptote is zero, in both the positive and negative direction.
Again, let's consult the factored form:
2 ÷ (x-3)(x+1)
There are vertical asymptotes at both x=3 and x=-1. As x approaches these numbers, depending on whether x is a little bigger or smaller than either one, y approaches positive and negative infinite, since the denominator of the function approaches zero.
Therefore, there are both positive and negative vertical asymptotes at both x=3 and x=-1
As for the graph, we'll leave that to you and the many applications that can aid in such a task!
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