Math 102, Chapter 4, Quiz

CHAPTER 4 “Quiz” Answers ( if you r questions aren’t here, they are very similar) 1. P(x) = 2x5 ! 2x4 ! 3x3 ! 6x2 ! 2x + 4 a. The degree of polynomial P is 5 5 and P must have zeros (roots). b. The y-intercept of the graph of P is (0,4) . The number of vertical asymptotes of P is c. According to Descartes’ Rule of Signs, P can have 0 or 2 0 . positive real zeros. There are two sign changes in P(x) = 2x5 ! 2x4 ! 3x3 ! 6x2 ! 2x + 4 . According to Descartes’ Rule of Signs, there can be either 2 or 0 positive roots. d. Similarly, P can have 1 or 3 negative real zeros. P cannot have 3 or 1 negative real roots because : P(–x) = 2(–x)5 ! 2(–x)4 ! 3(–x)3 ! 6(! x)2 ! 2(! x) + 4 = ! 2x5 ! 2x4 + 3x3 ! 6x2 + 2x + 4 ... three sign changes, so 3 or 1 positive roots for P(–x), so 3 or 1 negative roots for P(x). 2. List all theoretically possible rational roots of the polynomial: P(x) = 4x4 + 4x3 + 9x2+ 12x ! 9 Since this polynomial has all coefficients in the set of integers, any rational roots must be of the form p/q where p is a factor of the constant term (so p = ± 9, 3 or 1) and q is a factor of the leading coefficient (so q = ± 4, 2, or 1). Thus candidates for rational roots of this polynomial are: ±9 ±3 ±1 ±9/2 ± 3/2 ±1/2 ± 9/4 ± 3/4 ±1/4 (There are 18 in all.) 3. Construct the smallest polynomial with: real coefficients, roots ! 2 and 2 and 3 + i, with leading coefficient 5. You may leave the polynomial in factored form. Real coefficients + having root 3+i requires that it also have root 3! i. If “r” is a root, then (x ! r) must be a factor, so... P(x) = A (x ! ! 2) (x ! 2) (x ! (3+i) ) (x ! (3 ! i) ) 4. The polynomial P(x) = 2x4 ! 3x3 ! 7x2 ! 8x + 6 might have a zero at x= ½ or at x = 2. Use synthetic division to demonstrate that one of these IS, indeed, a zero, and the other is NOT. Identify which of these is a zero, and which is not. ½ 2 !3 2 1 !2 !7 !1 !8 !8 6 !4 !6 ! 12 0 2 2 !3 !7 2 4 1 2 !5 !8 6 ! 10 ! 9 ! 18 ! 3 Telling us that 2x4 ! 3x3 ! 7x2 ! 8x + 6 = (x ! ½ )(2x3 ! 2x2 ! 8x ! 12 Showing us that P(2) = ! 3 (So ½ is a zero of P .) (So 2 is NOT a zero of P .) 5. For each function below, list the equation(s) of the vertical and horizontal asymptote(s), if any. If there are none, write “none”. Vertical asymptote(s) Horizontal asymptote(s) 2 x !9 g(x) = ))))))) NONE y=1 x2 + 4 3x4 + x f(x) = )))))))) x = 3 and x = ! 3 NONE x2 ! 9 6x + 3 h(x) = ))))))))))) x = !1 y=0 2 x + 2x + 1 Vertical asymptotes of rational functions occur where the function grows unboundedly large ( or ) because the denominator shrinks toward 0 while the numerator does not shrink. This can only happen where the denominator of the rational function is 0. For g, the denominator, x2 + 4, is never 0, so g cannot have a vertical asymptote. For f, the denominator, x2 ! 9 = (x+3)(x! 3), is 0 at ! 3 and 3. The numerator is not 0 at x = ! 3 or 3. Therefore, it is clear that f grows unboundedly large as x approaches ! 3 and as x approaches 3. For h, the denominator is x 2 + 2x + 1 = (x+1)2 . This is 0 when x = ! 1. 6x+3 does not shrink toward 0 as x approaches ! 1. Thus there is a vertical asymptote at x = ! 1. Horizontal asymptotes of rational functions occur when the function values approach one particular number as x approaches infinity. This cannot occur when the degree of the numerator exceeds the degree of the denominator. So f has no horizontal asymptote (f resembles y = 3x2 when x is very large). g(x) approaches 1 as x ÷ 4 so g has HA y = 1. h(x) approaches 0 as x ÷ 4 so h has HA y = 0. 6. Sketch the graph of y = P(x) = 1 2 )) ( x + 2)2 (x — 3) Polynomial, domain is all of R, continuous, etc. . . . . . . . . . 18 . . . . . . . . . . . . . . . . . . . . . . . . . y-intercept: P(0) = ½ (0+2) (0 – 3) = –6 . . . . . . . . . . . . . . . . . x-intercept(s): P(x) = 0 . . . . . . . . . . . . . . . . . only when ( x + 2)2 (x – 3) = 0 . . . . . . . . . . . . . . . . . . . . when x = – 2 (twice), x = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . What happens to y=P(x) when x ÷ 4 . . . . . . . . . . . . . . . . . (increases towards 4) ? Consider that . . . . . . . . . . . . . . . . . (x + 2)2 and (x — 3) both grow unboundedly . . . . . . . . . . . . . . . . . -2 3 large, both positive, as x ÷ 4. So . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( ½ ) (x + 2)2 (x — 3) ÷ 4 as x ÷ 4. (i.e. grows unboundedly large) . . . . . . . . .— 6 . . . . . . . . . . . . . . . . . . . . . . . . . ... And when x ÷ -4, (x + 2)2 & (x — 3) both . . . . . . . . . . . . . . . . . grow unboundedly large, but (x—3) is large . . . . . . . . . . . . . . . . . negative, so the product is large & negative. . . . . . . . . . . . . . . . . . ( ½ ) (x + 2)2 (x — 3) ÷ -4 as x ÷ -4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Plotting a few additional points, such as . . . . . . . . . . . . . . . . . (–5,–36) & (–4,–14) & (1, –9) & (2, –8) & (4,18) adds accuracy & serves as a check on the analysis. 2 . . . . . . . . . . . . . . . . . . . . 7. 4 — 2x Sketch the graph of y = ))))))) Label all the intercepts & asymptotes. 3—x Rational function, therefore: Domain: y is defined for all values except where denominator = 0: x = 3 As x approaches 3, 4 — 2x ÿ –2, while 3 — x ÿ 0, so the quotient ÿ ± 4 Thus there is a vertical asymptote at x = 3.** y=2 As x ÿ 4 y ÿ 2 As x ÿ ! 4 y ÿ 2 Thus the function has a horizontal asymptoteu at y = 3. y-intercept: when x = 0 y = 4/3 x-intercept: when x = 2, y = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (0,4/3) . . . . . . (2,0) . . . . . . . . . . . . . . . . . . . . . x= 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ** More on the Vertical Asymptote: As x approaches 3 from below e.g. take x = 2.5 , y grows unboundedly large, negative: y = –1 , .5 2.8 , 2.9 , 2.99 , 2.9999 . . . ÷ 3 –1.6 , –1.8 , –1.98 , –1.9998 – .2 – .1 – .01 – .0001 ÷ 4 Similarly, computing function values for x that are just above 3, demonstrates that the function values are positive and grow unboundedly large (÷ 4) as x-values approach 3 from above. u Another way to discover the horizontal asymptote: ...divide & conquer: –2x + 4 –x+3 ))))))) = 2x – 4 x– 3 ))))))) = 2 2 + ))))))) x—3 & f(x) ÿ 2 + 0 As x ÿ as xÿ 4 4 , the latter portion ÿ 0