AnswersOC27 MATH COLL ALGEBRA-1200710-S1-Degele-VOL7006PGraphing Rational Functions

Graphs of Rational Functions Answers

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The graph of f(x) does not pass through (0, 0); therefore, f(x) has no real zeros.
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The graph of f(x) does not cross the x-axis; therefore, f(x) has no real zeros.
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The graph of f(x) does not cross the y-axis; therefore, f(x) has no real zeros.
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The graph of f(x) has one or more asymptotes; therefore, f(x) has no real zeros.
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The graph of has an oblique asymptote at y =

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x + 1.
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4x + 3.
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6x + 11.
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62x + 34.
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How would you describe the relationship between the real zero(s) and x-intercept(s) of the function

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When you set the function equal to zero, the solution is x = 1; therefore, the graph has an x-intercept of (1, 0).
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When you set the function equal to zero, the solutions are x = 0 or x = 1; therefore, the graph has x-intercepts at (0, 0) and (1, 0).
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When you substitute x = 0 into the function, there is no solution; therefore, the graph will not have any x-intercepts.
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Since there are asymptotes at x = –3, x = –1, and x = 0, the graph has no x-intercepts and, therefore, no real zeros.
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A plane travels 2,000 kilometers at a speed of 900 kilometers per hour (kph) with no wind. When a tailwind is present, the plane’s speed increases by x kilometers per hour. The time it takes the plane to travel the same distance with the tailwind, t(x), is defined by the function . What is the meaning of the y-intercept for this function?

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t
the time it takes the plane to travel without the tailwind
t
the speed of the plane when there is no tailwind present
t
the minimum amount of time it takes the plane to travel 2,000 km
t
the time it takes the plane to travel when the speed is decreased by 900 kph
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Which graph represents the rational function ?

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The graph of has a vertical asymptote at x = .

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Which graph represents the rational function ?

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f(x) has one real zero at –2 because the graph of the function has an intercept at (0, –2).
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f(x) has two real zeros at –4 and –2 because the graph of the function has intercepts at (–4, 0) and (0, –2).
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f(x) has no real zeros because the graph of the function does not pass through (0, 0).
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f(x) has one real zero at –4 because the graph of the function has an intercept at (–4, 0).
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The graph of has a horizontal asymptote at y = .

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