Explanation The function is h ( x ) = x − 2 x + 2 . The denominator of the given function becomes zero for x = − 2 where the function h is not defined. So, the domain of the function is ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) . Substitute 0 for x in the given function to find y -intercept. h ( 0 ) = 0 − 2 0 + 2 = − 1 Substitute 0 for h ( x ) in the given function to evaluate the x -intercept. x − 2 x + 2 = 0 x − 2 = 0 × ( x + 2 ) x − 2 = 0 x = 2 So, the x -intercept is ( 2 , 0 ) and y -intercept is ( 0 , − 1 ) . For a rational function f ( x ) = P ( x ) Q ( x ) ( P and Q are polynomial ) , a line x = a is vertical asymptote of the graph of f if Q ( x ) = 0 but P ( x ) ≠ 0 . For the function h ( x ) = x − 2 x + 2 , the denominator becomes zero at x = − 2 and at x = − 2 the numerator becomes, P ( x ) = − 2 − 2 = − 4 ≠ 0 So, x = − 2 is the vertical asymptote of h ( x ) = x − 2 x + 2 . For a rational function f ( x ) = P ( x ) Q ( x ) ( P and Q are polynomial ) , a line y = b is horizontal asymptote of the graph of f if either lim x → ∞ f ( x ) = b or lim x → − ∞ f ( x ) = b . Take lim x → ∞ both sides of the function h ( x ) = x − 2 x + 2 . Then, the above expression becomes, lim x → ∞ h ( x ) = lim x → ∞ ( x − 2 x + 2 ) = lim x → ∞ ( 1 − 2 x 1 + 2 x ) ( Take x as common from Numerator and Denominator ) = 1 So, y = 1 is horizontal asymptote of the function h ( x ) = x − 2 x + 2 . In order to find the critical point first differentiate the function and then equate to zero

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 4, Problem 17RE

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To sketch: The graph of the given function.

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Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The graph of the given function.

Solution Summary: The author illustrates the graph of a given function. The denominator becomes zero at x=-2 where the function h is not defined.

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