Solving Rational Expressions
Graphing Rational Functions
Analyzing Rational Functions
100
Solve Algebraically x - (10/x) = 3
What is x = -2 or x = 5
100
Graph y = - 1/(x-2). State the domain, range, y-intercept, and identify any asymptotes.
What is Domain: x < 2 or x > 2 Range: y < 0 or y > 0 y-intercept: (0, 1/2) The graph has a vertical asymptote at x = 2 and a horizontal asymptote at y = 0. *See Doc in Google Classroom for Graphs.
100
Describe how the graph of g(x) is related to the graph of ƒ(x) = 1/x if g(x) = 1/(x+4)
What is g is f translated 4 units left
200
Solve Algebraically 5/(x+1) = 2/(x+4)
What is x = -6
200
Graph using long division to rewrite the function in the form g(x) = a[1/(x - h) ]+k or g(x) = (1/[(1/b)(x - h)]) + k (3x - 4)/(x - 1)
What is See Doc in Google Classroom for image. The graph of g (x) has vertical asymptote x = 1, horizontal asymptote y = 3, and reference points (-1+1, -(-1)+3) = (0,4) and (1+1, -(1)+3) = (2,2) . Domain of g (x): Range of g (x): Inequality: x < 1 or x > 1 Inequality: y < 3 or y > 3
200
Describe how the graph of g(x) is related to the graph of ƒ(x) = 1/x if g(x) = [1/(x-2)]+ 3
What is g is f translated 2 units right and 3 units up
300
Solve algebraically 3/(x+2) + 3/(2x+4) = x/(2x+4)
What is x = 9
300
Graph y = (2x^2)/(x+2). State the domain, range, x-intercept, and identify any asymptotes.
What is Domain: x < -2 or x > -2 Range: y ≤ -16 or y ≥ 0 The numerator has 0 as its only zero, so the x-intercept is (0, 0). The denominator has -2 as its only zero, so the graph has a vertical asymptote at x = -2.
300
State the domain, x-intercept(s), and identify asymptotes of ƒ(x) = (x^2-3x)/(x+4)
What is Domain: x < -4 or x > -4 x-intercepts: x = 0, 3 asymptotes: x = -4
400
Solve x/(x-3) + x/2 = 6x/(2x-6)
What is x = 0 or x = 7
400
Graph ƒ(x) = (x+1)/(x-2) Also state the function’s domain and range.
What is The function is undefined when x - 2 = 0, or x = 2. Since x - 2 is not a factor of the numerator, there is a vertical asymptote rather than a “hole” at x = 2. The numerator and denominator have the same degree and the leading coefficient of each is 1, so there is a horizontal asymptote at y = 1. An x-intercept occurs when x + 1 = 0, or x = -1. Domain: Inequality: x < 2 or x > 2 Range: Inequality: y < 1 or y > 1 See Google Doc for graph.
400
State the domain, x-intercept(s), and identify asymptotes of ƒ(x) = (x-3) / (x^2+6x+5)
What is Domain: x<-5 or -5-1 x-intercept: x = 3 asymptotes: x = -5, x = -1, y = 0
500
Solve 3x/(x+1) + 6/2x = 7/x
What is x = -2/3 x = 2
500
ƒ(x) = (x+1)(x-1) / (x+2)
What is There is a vertical asymptote at x = -2. f(x) = (x + 1)(x - 1) /(x+2) = (x^2-1)/(x+2) The line y = x - 2 is a slant asymptote. There are x-intercepts at x = -1 and x = 1. Check the sign of the function on the intervals x < -2, -2 < x < -1, -1 < x < 1, and x > 1. Domain: Inequality: x < -2 or x > -2 Range (approximate): Inequality: y< -7.46 or y > -0.54
500
How do you identify any linear asymptotes of a rational function?
What is The graph has vertical asymptotes for every value of the variable that makes the denominator 0 but does not make the numerator 0. It has a horizontal asymptote if the degree of the denominator is less than or equal to the degree of the numerator. The graph has a slant asymptote if the degree of the numerator is 1 more than the degree of the denominator.
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