SOLUTION: Let 'f' be the function whose graph is obtained by translating the graph of y=1/x to the right 3 units and upward 2 units.
(a) Write an equation for f(x) as a quotient of two poly
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-> SOLUTION: Let 'f' be the function whose graph is obtained by translating the graph of y=1/x to the right 3 units and upward 2 units.
(a) Write an equation for f(x) as a quotient of two poly
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Question 117539: Let 'f' be the function whose graph is obtained by translating the graph of y=1/x to the right 3 units and upward 2 units.
(a) Write an equation for f(x) as a quotient of two polynomials.
(b) determind the zeros of 'f'.
(c) identify the asymptotes of the graph of f(x). Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website! To move a function up, just subtract the amount OUTSIDE of the function. If I want to move h(x) up 2 units to make f(x), then f(x) = h(x) + 3. To move a function to the right subtract the amount INSIDE the function. If I want to move h(x) to the right 3 units, then f(x) = h(x - 3).
So the f(x) that we are looking for, given , is .
Now we need this to be in the form of the quotient of two polynomials, so let's solve for y and simplify:
Vertical Asymptotes.
Look for values of x not in the domain of the function. In this case the domain of the function is all reals except x = 3, therefore there is one vertical asymptote at x = 3.
Horizontal Asymptote.
Since the degree of the numerator polynomial and the denominator polynomial are equal, there is a horizontal asymptote at where a is the lead coefficient on the numerator and b is the lead coefficient on the denominator. In this case,
There is no slant asymptote. Slant asymptotes only occur where the numerator polynomial is of larger degree than the denominator polynomial.