Solutions to Extra Practice for Chapter 4

Solutions to Extra Practice for Chapter 4 1. Graph a polynomial function. Label all intercepts and describe the end behavior. a. P(x) = x4 ! 2x3 ! 15x2. (1) (2) (3) Domain = R, of course (since this is a polynomial function... domain must be R.) y-intercept: P(0) = 04 ! 2C03 ! 15C02 = 0 x-intercepts: x4 ! 2x3 ! 15x2 = 0 x2 ( x2 ! 2x ! 15) = 0 x2 ( x ! 5) (x + 3) = 0 x = 0 or 0 or 5 or ! 3 (4) the outer limits: As x 64, x4 dominates the other terms, so f(x) 64. (i.e. f(x) grows unboundedly large) Likewise, as x 64, x4, dominating the other terms, takes f(x) 64 (5) ...the graph: !3 0 5 !2 !1 0 1 ! 100 b. (1) (2) (3) P(x) = x5 ! 5x3 + 4x. Domain = R y-intercept: P(0) = 0 x-intercepts : x5 ! 5x3 + 4x = 0 when x(x4 ! 5x2 + 4) = 0 x(x2 ! 4)(x2 ! 1) = 0 x(x+2)(x! 2)(x+1)(x! 1) = 0 x = 0,! 2, 2, ! 1, 1 (4) outer limits: (5) see above, right. P is an odd function. as x64, f(x), dominated by x5, 64 ax x6!4 , f(x), again ruled by x5, 6!4 Note the curve is smoother than this pathetic illustration indicates. It is just difficult to draw these curves with a mouse and flaky software. After the seventeenth try, patience vanishes. 2 2. Use polynomial long division. EG: Divide ... 6x4 + 2x2 + 22x 2x + 5 ))))))))))))))) 2 2x2 + 5 3x2 ! 13/  6x + 0 + 2x + 22x + 0 6x4 +15x2 )))))))))))))2 ! 13x ! 65/ ! 13x2 ))))))))))))) 265 22x + /2 2 )))))))))))))))))))))))) 2 4 SO 6x4 + 2x2 + 22x ))))))))))))))) = 3x2 ! 13/2 + 2x2 + 5 OR 6x4 + 2x2 + 22x = 22x + 65/2 ))))))))) 2x2 + 5 (3x2 ! 13/2) ( 2x2 + 5) + 22x + 65/2 Notice the last statement shows the way to check your division results!! 3. a. The graph shows us there is a repeated zero*, of even order, at x = 0, and a simple zero at x = 3. It also tells us P(2) must be 8. The zeroes inform us that (x-0)(x-0)(x-3) must be part of the factorization of P. Therefor, P(x) = xCxC(x! 3)C(other factors). The degree of P , which is 3, tells us that the “other factors” can be only constant. Stated briefly, we know: Roots are 0, 0, and 3 & polynomial has degree 3, so P(x) = K x2 (x! 3) Since P(2) = 8: 8 = P(2) = KC22(2! 3) = -4K, so K must be ! 2. Therefor, P(x) = ! 2 x2 (x! 3) , or P(x) = ! 2x3 + 6 x2 b. Find a second-degree polynomial with real coefficients with roots ! 3 and 2, passing through (4,3). r is a root of a polynomial P if, and only if, (x ! r) is a factor of P(x). Having roots ! 3 and 2 requires that both (x ! ! 3) and (x ! 2) be factors. Since P must be a 2nd-degree polynomial, the only other possible factor is a constant. So: P(x) = A (x + 3) (x ! 2) where A is some currently undetermined constant. The additional requirement that the curve contain (4,3) determines A: P(4) = A (4+3) (4 ! 2) = A (7) (2) = 14A But P(4) must be 3... so 14A must be 3. therefore A = 3/14 and P(x) = (3/14) (x + 3) (x ! 2) = (3/14) ( x2 + x ! 6) c. Find a fourth degree polynomial with real coefficients, with roots 2i and 1 ! i, and constant term 16. The complex roots of a polynomial with REAL coefficients must occur in conjugate pairs. Since we are required to find a polynomial with real coefficients, and since 1 ! i is a root, its conjugate, 1 + i must also be a root. Likewise, since 2i is a root, ! 2i must also be a root. Therefore: P(x) = A ( x ! (1 + i) ) ( x ! (1 ! i) ) ( x ! 2i ) ( x + 2i ) = A ( x2 ! ( 1+i + 1! i ) x + (1+i)(1! i) ) ( x2 ! 4i2 ) = A ( x2 ! 4. 2x + ) ( x2 + 4 ) 2 The additional requirement that the constant term be 16 determines A. A(2)(4) must be 16.... thus A must be 2, and P(x) = 2 (x 2 ! 2x+ 2) ( x2 + 4 ) = 2 (x4 ! 2x3+ 6x2 ! 8x + 8) = 2x4 ! 4x3+ 12x2 ! 16x + 16 Find all the zeros of a polynomial function. a. For P(x) = 5x3 ! 22x2 + 18x ! 4, list all the theoretically possible rational roots of P; and use synthetic division to locate one; then find the remaining roots. RATIONAL roots of a polynomial with INTEGER coefficients must be of the form p/q where p divides the constant term and q divides the leading coefficient. So any rational roots of 5x3 ! 22x2 + 18x ! 4 must be of the form: ±4 or ±2 or ± 1 1 2 4 ))))))))))))))) which yields: ±1 ±2 ±4 ±) ±) ±) ±5 or ±1 5 5 5 TESTING we divide 5x3 ! 22x2 + 18x ! 4 by x ! r for each potential root r: 1 !2 1 /5 5 !22 18 !4 5 !17 1 5 !17 1 !3 2 5 !22 18 !4 10 !24 !12 5 !12 -6 ; 4 5 !22 18 !4 20 !8 40 5 !2 10 ; !4 5 !22 18 !4 !10 128 !12 5 !32 ; !1 /5 5 !22 18 !4 !1 ! 2 3 / 5 ; 5 !23 6 7 / 5 ; 2 5 !22 18 !4 2 !8 4 5 !20 10 0 ; !4 /2 5 69 So P(x) = (x ! 5 !22 18 !4 !5 27 5 !27 ; ; 5 !22 18 !4 !10 !17 1 5 !32 1 !3 5 !22 18 1 !2 1 /5 5 !21 6 9 / 5 !1 ; 2 2 /5 )(5x /5 ! 20x + 10) = (5x ! 2)(x2 ! 4x + 2) finally !** For the remaining roots we solve: x2 ! 4x + 2 = 0 ... using the quadratic formula: x = 4 ± q%%1%%6%%%%!%%8% = 2 ± q%%2 ))))))))))) 2 PS P(-x) = ! 5x3 ! 22x2 ! 18x ! 4, so, by Descartes’ Rule of Signs, there are no negative real roots. (We could have saved time **here!!!) b. Find all the roots of P(x) = 4x5 + 15x3 ! 4x 4x5 + 15x3 ! 4x = x(4x4 + 15x2 ! 4) = x (4x2 ! 1)(x2 + 4) = x(2x+1)(2x! 1)(x +2i)(x - 2i) Notice you can factor this! ...and this! So the roots of P are: 0, ! ½ , ½ , ! 2i , 2i 5. a. 3x ! 4 f(x) = ))))) x!2 x=2 C Find the x and y intercepts. f(0) = 2; f(x) = 0 when 3x ! 4 = 0 ... x = 4/3 C Find the equation of the vertical asymptote. Near x = 2, numerator 6 2, denominator6 0, so fraction 6 ±4 ... vertical asymptote at x = 2. C Find the equation of the horizontal asymptote. As x 64, (3x ! 4)/(x ! 2) 6 3 so y = 3 is the horizontal asymptote. C Division also shows us: 3x ! 4 2 ))))) = 3 + )))) , which would have to be ... x!2 x!2 y=1/x, shifted 2 units right, stretched vertically by factor 2, then shifted upwards 3 units. b. x2 ! 3x ! 4 f(x) = )))))))))) = x!2 y=3 (0,2) (4/3,0) (curve should have no “squiggles”.) (x ! 4)(x + 1) )))))))))) x!2 x=2 C Intercepts: f(0) = 2. f(x) = 0 when x = 4, ! 1. C Vertical asymptote At x = 2, for the usual reason. C Horizontal asymptote None. Divide to see other asymptote: !6 x2 ! 3x ! 4 ))))))))) = x ! 1 + )))) x!2 x!2 So we can see that y = x ! 1 is an oblique linear asymptote for y = f(x). (0,2) -1 y=x!1 4 (x ! 1)(x2 + x + 1) x3 ! 1 f(x) = ))))) = )))))))))))))) x x c. x-intercept (1,0)** vertical asymptote at x = 0 x3 ! 1 f(x) = ))))) = x2 ! )) x x 1 y = x2 ... this shows the given f is x 2 + an amount that is very small when x is very large. So when x is large, the graph of y = f(x) is very close to the graph of the familiar curve y = x 2. And as x÷ 4, the difference ÷ zero. This is asymptotic behavior. (1,0) x=0 NO worries: On the test, the only asymptotes you will be asked to find are linear asymptotes!!! **of course, we determined that the remaining roots are complex: x2 + x + 1 = 0 x= 6. ! 1± q%%1%%!%%4% = ! 1 ± ))))))))) ) 2 2 3i 2 q%% ))) The Remainder Theorem gives us the answers: If polynomial P(x) = (x ! c) Q(x) + R(x) where Q & R are the quotient and remainder polynomials for P ÷ ( x! c ), then R(x) is a constant, and P(c) = that constant. a. Find the value of P(x) = 2x5 ! 20x4 ! 20x3 ! 20 x2 ! 20x ! 22 @ 11 — without raising 11 to a power. 11 2 —20 —20 —20 —20 —22 22 22 22 22 22 2 2 2 2 2 0 This tells us that P(11) = 0, thus 11 is a root, and (x — 11) is a factor of P. In fact, this tells us: P(x) = (x — 11) 2 ( x4 + x 3 + x 4 + x 3 + x + 1 ) b. If P(x) = (x! 2)(x3 ! 4x2 + 7x + 13) + 7 , what is P(2)? P(2) = (2 ! 2) ( whatever) =0 +7 +7 SO: P(2) = 7 ... JUST as the remainder theorem says.