Question

A piecewise function f (x) is defined by Part A: Graph the piecewise function f (x) and determine the range. (5 points)Part B: Determine the asymptotes of f (x). Show all necessary calculations. (5 points)Part C: Describe the end behavior of f (x). (5 points)

Answer 1

• a) Range: (-∞, –1)

• b) Horizontal asymptote when x ≤ 2: y = –3. Vertical asymptote when x > 2: x= 2. Horizontal asymptote when x > 2: y = –1.

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• c) When x decreases, it approaches –3, and when x increases, it approaches –1.

Step-by-step explanation

• Part A: Graph the piecewise function f (x) and determine the range.

Using a graphing tool we can get the following graph:

As the range is all the y-values included in the function, we can see that in this case, the range is:

`(-\infty,-1)`

• Part B: Determine the asymptotes of f (x). Show all necessary calculations.

We have to get the asymptotes from the part where x ≤ 2 and x > 2.

In x ≤ 2 we do not have a vertical asymptote as it is not a rational function, but we do have a horizontal asymptote, which is y = k when we have the function:

`y=a(b)^(x-h)+k`

Thus, in our case y = –3.

In x > 2, we do have a vertical and horizontal asymptote. We can get the vertical asymptote by setting the denominator of the function to 0:

`x^2-5x+6=0`

If we solve the expression by factoring we get two solutions:

`(x-2)(x-3)=0`
`\begin{gathered} x_1-2=0 \\ x_1=2 \end{gathered}`
`\begin{gathered} x_2-3=0 \\ x_2=3 \end{gathered}`

Based on our function we can see that the vertical asymptote is x = 2 when x > 2. Then, the horizontal asymptote can be calculated with the limit:

`\lim_(x\to\infty)((-x^2+2x+3)/(x^2-5x+6))=-1`

Thus, our horizontal asymptote is y = –1 when x > 2.

• Part C: Describe the end behavior of f (x).

The end behavior of a function is how the function acts when x increases or decreases. Based on our graph we can see that when x decreases, it approaches –3, and when x increases, it approaches –1.