1.6 Miscellaneous Functions

Answers and Explanations

Parametric Equations

* 1.    (B) Graph these parametric equations for values of between –5 and 5 and for and between –2.5 and 2.5.

            Apparently the values are always greater than some value. Use the TRACE function to move the cursor as far left on the graph as it will go. This leads to a (correct) guess of . This can be verified by completing the square on the equation: 

            This represents a parabola that opens up with vertex at . Therefore, .

2.     (D) D is the only reasonable answer choice. To verify this, note that . So . Adding this to = sin2gives . Since 0   1 because 0  sin2  1, this can only be a portion of the parabola given by the equation y2 + 4= 4.

* 3.     (A) You could graph all three parametric pairs to discover that only I gives a circle. (II and III give semicircles). You can also see this by a simple analysis of the equations. Removing the parameter in I by squaring and adding gives x2 + y2 = 1, which is a circle of radius 1. Substitutingfor in the equation of II and squaring gives x2 + y2 = 1, but ≥ 0 so this is only a semicircle. Squaring and substituting x2 for in the equation of III gives x2 + y2 = 1, but ≥ 0 and so this is only a semicircle.

Piecewise Functions

* 1.     (B) Enter abs(2– 1) into Y1 and 4+ 5 into Y2. It is clear from the standard window that the two graphs intersect only at one point.

* 2.     (E) Enter abs(– 2) into Y1, 1 into Y2, and 4 into Y3. An inspection of the graphs shows that the values of for which the graph of Y1 is between the other two graphs are in two intervals. E is the only answer choice having this configuration.

* 3.     (A) Subtract |x| from both sides of the equation. Since |y| cannot be negative, graph the piecewise function

            In the first command, the word “and” is in TEST/LOGIC. The result is a square that is  on a side. Therefore, the area is 8.

4.     (A) Since f(x) must = f(|x|), the graph must be symmetric about the y-axis. The only graph meeting this requirement is Choice A.

* 5.        (B) Since the point where a major change takes place is at (1,1), the expression in the absolute value should equal zero when = 1. This occurs only in Choice B. Check your answer by graphing the function in B on your graphing calculator.

6.     (E) Choice A fails if = 0.5. Choice B subtracts cents from ounces. Choice C fails if = 1. Choice D adds cents to ounces.

7.     (C) Since f(x) = an integer by definition, the answer is Choice C.

* 8.        (E) Enter int(4x) – 2into Y1. The graph is shown in the figure below.

            The breaks in the graph indicate that it cannot be the graph of any of the first four answer choices.