1. Radicals 2.6 There are answers to some textbook questions and then there is an introduction to finding slant asymptotes.

2. Page 193 1.Domain: VA : x = 1 HA : y = 0 2.Domain: VA : x = 1 HA : y = 5 3.Domain: VA : x = 1, x = -1 HA : y = 3 4.Domain: VA : x = 1, x = -1 HA : y = 0 5.Domain: VA : x = 0, HA : y = 0 7.Domain: VA : x = 2, HA : y = 1 8. Domain: VA : x = -1/2, HA : y = -5/2 11. Domain: VA : none, HA : y = 3 12. Same answers as #11 13. d 14. a 15. c 16. b 17.1 18. none 19. 6 20. 2 27.Domain: VA: x = - 2 HA: y = 0 y intercept: ( 0, ½) 28.Domain: VA: x = 3 HA: y = 0 y intercept: ( 0, -1/3) 1.

3. Page 193 29. Domain: VA : x = -2 HA : y = -1 y int: ( 0, -1/2) 30. Domain: VA : x = 3 HA : y = 0 y int: ( 0, -1/3) 31. Domain: VA : x = -1, HA : y =2 y int: (0, 5) 32. Domain: VA: x = - 1 HA: y = 3 y int: ( 0, 1) 33. Domain: VA: none HA: y = 1 y int: (0,0) 34. Domain: VA: x = 0 HA: y = -2 y int: none t int: (1/2, 0)

4. If…………. Degree of Numerator Degree of Denominator then HA is the graph of the polynomial resulting from long division.

5. CONSIDER… Note: that degree of numerator is > than degree of denominator Means we MUST use long division to determine the asymptote. (slant asymptote)

6. The slant asymptote is the graph of the line: y = x – 2. (Ignore the remainder because as x gets bigger and bigger that fraction gets closer and closer to zero.) This function has TWO asymptotes: a vertical asymptote of x = -1 and a slant asymptote of y = x - 2

7. CONSIDER… Factoring the numerator will make it easier to find the x intercepts: (x – 2)(x + 1) or (2,0) and (-1,0) and the y intercept is (0,2) The vertical asymptote is x = 1.

8. The slant asymptote is the graph of the line: y = x y intercept is (0, 2)

9. Radicals 2.6

10. Whereas vertical asymptotes are sacred ground, horizontal asymptotes are just useful suggestions. Whereas you can never touch a vertical asymptote, you can (and often do) touch and even cross horizontal asymptotes. Whereas vertical asymptotes indicate very specific behavior (on the graph), usually close to the origin, horizontal asymptotes indicate general behavior far off to the sides of the graph. To get the idea of horizontal asymptotes, let's looks at some simple examples. Find the horizontal asymptote of the following function: The horizontal asymptote tells me, roughly, where the graph will go when x is really, really big. So I'll look at some very big values for x, some values of x very far from the origin: x – 100 000 – 0.0000099... – 10 000 – 0.0000999... – 1 000 – 0.0009979... – 100 – 0.0097990... – 10 – 0.0792079... –1–1 0.5 02 11.5 100.1188118... 1000.0101989... 1 0000.0010019... 10 0000.0001000... 100 0000.0000100... Off to the sides of the graph, where x is strongly negative (such as – 1,000 ) or strongly positive (such as 10000 ), the " +2 " and the " +1 " in the expression for y really don't matter so much. I ended up having a really big number divided by a really big number squared, which "simplified" to be a very small number. The y -value came mostly from the " x " and the " x 2 ". And since the x 2 was "bigger" than the x, the x 2 dragged the whole fraction down to y = 0 (that is, the x -axis) when x got big. I can see this behavior on the graph: The graph shows some slightly interesting behavior in the middle, near the origin, but the rest of the graph is fairly boring, trailing along the x -axis. If I zoom in on the origin, I can also see that the graph crosses the horizontal asymptote (at the arrow): Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved (It is common and perfectly okay to cross a horizontal asymptote. It's the vertical asymptotes that I'm not allowed to touch.) As I can see in the table of values and the graph, the horizontal asymptote is the x -axis. horizontal asymptote: y = 0 (the x -axis) In the above exercise, the degree on the denominator (namely, 2 ) was bigger than the degree on the numerator (namely, 1 ), and the horizontal asymptote was y = 0 (the x -axis). This property is always true: If the degree on x in the denominator is larger than the degree on x in the numerator, then the denominator, being "stronger", pulls the fraction down to the x -axis when x gets big. That is, if the polynomial in the denominator has a bigger leading exponent than the polynomial in the numerator, then the graph trails along the x -axis at the far right and the far left of the graph. x – 100 000 1.9999999... – 10 000 1.9999997... – 1 000 1.9999710... – 100 1.9971026... – 10 1.7339449... –1–1 –0.9 0 – 1.2222222... 1 –0.9 10 1.7339449... 100 1.9971026... 1 000 1.9999710... 10 000 1.9999997... 100 000 1.9999999... For big values of x, the graph is, as expected, very close to y = 2. The graph reflects this: Then my answer is: horizontal asymptote: y = 2 In the example above, the degrees on the numerator and denominator were the same, and the horizontal asymptote turned out to be the horizontal line whose y -value was equal to the value found by dividing the leading coefficients of the two polynomials. This is always true: When the degrees of the numerator and the denominator are the same, then the horizontal asymptote is found by dividing the leading terms, so the asymptote is given by: y = (numerator's leading coefficient) / (denominator's leading coefficient)

11. NONLINEAR INEQUALITIES y intercept is (0, 2) 2013 STUDENTS: THIS IS NOT ON THIS TEST!!!!!!!!!!!!!!!

12. Factor: (x – 3)(x + 2) This gives us the critical numbers which are x = -2 and x = 3. These numbers set the parameters of our test intervals. We now find an x-value within these three test areas and evaluate the function at these three points to determine if the function is positive or negative at those points. Let x = -3 (-3) 2 – (-3) – 6 results in 6 which is POSITIVE Let x = 0 (0) 2 – (0) – 6 results in -6 which is NEGATIVE Let x = 4 (4) 2 – (4) – 6 results in 6 which is POSITIVE WE CONCLUDE THAT THE FUNCTION IS NOT SATISFIED WHEN THE RESULT IS “ NEGATIVE ”.

13. SOLVE P/Q SYN DIV Factor by group This makes the critical numbers -4, 3/2 and 4.