SOLUTION: How and what is the answer of the following?
Mrs.cruz asks the class to graph the functions f(x) = x-2/x and g(x) = 3-2/x. Arnold and beth are asked to go to the chalkboard to gr
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-> SOLUTION: How and what is the answer of the following?
Mrs.cruz asks the class to graph the functions f(x) = x-2/x and g(x) = 3-2/x. Arnold and beth are asked to go to the chalkboard to gr
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Question 1167698: How and what is the answer of the following?
Mrs.cruz asks the class to graph the functions f(x) = x-2/x and g(x) = 3-2/x. Arnold and beth are asked to go to the chalkboard to graph f(x) = x-2/x and g(x) = 3- 2/x respectively.
a.Mrs.cruz asks Arnold and Beth to first writethedomain of each function. What would arnold write? What would beth write?
b.Mrs.cruz asks both students to determine the asymptotes of each function. What should each student say? Answer by Theo(13342) (Show Source):
place all terms under a common denominator to get:
y = (x^2 - 2)/x
the vertical asymptote is at x = 0.
since the degree of the numerator is 1 more than the degree of the denominator, the asymptote will be found by dividing the numerator by the denominator and using the quotient in the equation of the asymptote.
y = (x^2 - x) / x = x with a remainder of -2.
the quotient is x and the remainder is -2.
the equation of the quotient is y = x
the domain of the function is all values of x except at x = 0.
the range of the function is all values of y.
as x approaches 0 from the left, y approaches minus infinity.
as x approaches 0 from the right, y approaches plus infinity.
the slant asymptote is the line y = x.
this is what the graph looks like.
beth's equation is:
g(x) = 3 - 2/x
replace g(x) with y to get:
y = 3 - 2/x
place all terms under a common denominator to get:
y = (3x - 2) / x
the degree of the numerator is the same degree as the denominator, so there will be a horizontal asymptote.
divide 3x - 2 by x to get a quotient of 3 with a remainder of 2.
the equation of the asymptote will be y = 3
this is what the graph looks like.
both equations have a vertical asymptote at x = 0
therefore, the domain of both equations is all values of x except at x = 0.
the range of both functions is equal to all real values of y.
the fact that y is undefined at x = 0 doesn't affect the range of the equations.
there are no values of y that are not in the range of y .
in addition to the vertical asymptotes, .....
arnold's equation has a slant asymptote at y = x.
beth's equation has a horizontal asymptote at y = 3
here's a reference on horizontal and slant asymptotes.
