Question

(01.03. 106, 1.08 HC)

Answer 1

Answer:

  A) see attached for a graph. Range: (-∞, 7]

  B) asymptotes: x = 1, y = -2, y = -1

  C) (x → -∞, y → -2), (x → ∞, y → -1)

Step-by-step explanation:

Part A

A graphing calculator is useful for graphing the function. We note that the part for x > 1 can be simplified:

  $\dfrac{-x^2+x+2}{x^2-3x+2}=-\dfrac{(x-2)(x+1)}{(x-2)(x-1)}=-\dfrac{x+1}{x-1}\quad x\ne 2$

This has a vertical asymptote at x=1, and a hole at x=2.

The function for x ≤ 1 is an ordinary exponential function, shifted left 1 unit and down 2 units. Its maximum value of 3^-2 = 7 is found at x=1.

The graph is attached.

The range of the function is (-∞, 7].

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Part B

As we mentioned in Part A, there is a vertical asymptote at x = 1. This is where the denominator (x-1) is zero.

The exponential function has a horizontal asymptote of y = -2; the rational function has a horizontal asymptote of y = (-x/x) = -1. The horizontal asymptote of the exponential would ordinarily be y=0, but this function has been translated down 2 units.

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Part C

The end behavior is defined by the horizontal asymptotes:

  for x → -∞, y → -2

  for x → ∞, y → -1