Extra Polynomial & Rational Practice!
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1 Extra Polynomial & Rational Practice! EPRP- p1 1. Graph these polynomial functions. Label all intercepts and describe the end behavior. 3 a. P(x = x x 1x. b. P(x = x x x.. Use polynomial long division. EG: Divide... 6x x x x (, (, 1 (3, 3. Construct a polynomial to specifications! a. Find a second-degree polynomial with real coefficients, with zeroes 3 and, passing through (,3. b. Find a polynomial whose graph is be similar to that shown here. c. Find a polynomial of minimal degree, with real coefficients, having zeroes i and 1 i, and constant term 16.. Find all the zeros of these polynomial functions. 3 a. For P(x = x x 18x, list all the theoretically possible rational zeroes of P; use synthetic division to locate a zero; then find the remaining zeroes. b. Find all the zeroes of P(x = x 1x x. Graph these rational functions. Include the following: Find the x and y intercepts. Find the equation of any vertical asymptote. Find the equation of any horizontal asymptote. Sketch the graph. 3x a. f(x = x x 3x b. f(x = x 3 x 1 c. f(x = x x x 1 d. f(x = x 6. Miscellaneous: esca - p 3 a. Find the value of P(x = x x x x x without raising 11 to a power. 3 b. If P(x = (x (x x 7x 13 7, what is P(?
2 Solutions to Extra Polynomial & Rational Practice EPRP- p 1. Graph a polynomial function. Label all intercepts and describe the end behavior. 3 a. P(x = x x 1x. (1 Domain = R, of course (since this is a polynomial function... domain must be R. 3 ( y-intercept: P( = 1 = 3 (3 x-intercepts: x x 1x = SIGNS: P(x is x ( x x 1 = when x is between 3 x ( x (x 3 = x = or or or 3 ( the outer limits: As x, x dominates the other terms, so f(x. (i.e. f(x grows unboundedly large Likewise, as x, x, dominating the other terms, takes f(x (...the graph: b. P(x = x x x. (1 Domain = R ( y-intercept: when x =, y = P( = (3 x-intercepts: y = when x x x = SIGNS: P(x is when x (x x = when x is between -1 1 x (x (x 1 = x (x(x (x1(x 1 = x =,,, 1, 1 ( the outer limits: as x, f(x, dominated by x, ax x, f(x, again ruled by x, ( graph above, right. P is an odd function. 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. When I win the lottery, I will buy some fancy software. esca - p
3 . Use polynomial long division. 6x x x EG: Divide... x 13 3x / x 6x x x 6x 1x 13x 6 13x / 6 x / 6 SO 6x x x 13 x / = 3x / x x EPRP- p OR 6x x x = (3x / ( x x / Notice the last statement shows the way to check your division results ( multiply & add!! 3. a. Find a second-degree polynomial with real coefficients with zeroes 3 and, passing through (,3. r is a zero of polynomial P if, and only if, (x r is a factor of P(x. Having zeroes 3 and requires that both (x 3 and (x be factors. nd Since P must be a -degree polynomial, the only other possible factor is a constant. So: P(x = A (x 3 (x where A is some currently undetermined constant. The additional requirement that the curve contain (,3 determines A: P( = A (3 ( = A (7 ( = 1A But P( must be 3... so 1A must be 3. therefore A = 3/1 and P(x = (3/1 (x 3 (x = (3/1 ( x x 6 b. The graph shows us there is a repeated zero*, of even order, at x =, and a simple zero at x = 3. It also tells us P( must be 8. The zeroes inform us that (x-(x-(x-3 must be part of the factorization of P. Therefor, P(x = x x (x 3 (other factors. The degree of P, which is 3, tells us that the other factors can be only constant. Stated briefly, we know: Zeroes are,, and 3 & polynomial has degree 3, so P(x = K x (x 3 Since P( = 8: 8 = P( = K ( 3 = -K, so K must be. 3 Therefore, P(x = x (x 3, or P(x = x 6 x esca csts - p3
4 EPRP- p c. Find a fourth degree polynomial with real coefficients, with zeroes i and 1 I, and constant term 16. The complex zeroes 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 zero, its conjugate, 1 I must also be a zero. (Otherwise we d end up with complex coefficients. Likewise, since i is a zero, i must also be a zero. Therefore: P(x = A ( x (1 i ( x (1 I ( x i ( x i = A ( x ( = A ( x 1i 1 I x (1i(1 i ( x i x ( x The additional requirement that the constant term be 16 determines A. A(( must be thus A must be, and P(x = (x x ( x 3 = (x x 6x 8x 8 3. Find all the zeros of a polynomial function. = x x 1x 16x 16 3 a. For P(x = x x 18x, list all the theoretically possible rational zeroes of P; and use synthetic division to locate one; then find the remaining zeroes. RATIONAL zeroes 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. 3 So any rational zeroes of x x 18x must be of the form: ± or ± or ± 1 1 which yields: ±1 ± ± ± ± ± ± or ±1 3 TESTING we divide x x 18x by x r for each potential zero r: / 18 / 18 / 18 1 / / 1 / / 3 67 / So P(x = (x / (x x 1 = (x (x x For the remaining zeroes we solve: x x =... using the quadratic formula: x = ± = ± finally!** 3 PS P(-x = x x 18x, so, by Descartes Rule of Signs, there are no negative real zeroes. (We could have saved time **here!!! b. Find all the zeroes of P(x = x 1x x esca csts - p x 1x x = x(x 1x = x (x 1(x = x(x1(x 1(x i(x - i Notice you can factor this!...and this! So the zeroes of P are:, ½, ½, i, i
5 EPRP- p 3x. a. f(x = x x = Find the x and y intercepts. f( = ; f(x = when 3x =... x = / 3 Find the equation of the vertical asymptote. y = 3 Near x =, numerator, denominator, so fraction ±... vertical asymptote at x =. (, Find the equation of the horizontal asymptote. As x, (3x /(x 3 so y = 3 is the horizontal asymptote. Division also shows us: 3x = 3, which would have to be... x x y=1/x, shifted units right, stretched vertically by factor, then shifted upwards 3 units. (In reality, curve has no squiggles. Mice!!!! SIGNS: f(x is when x is between: / 3 (/3, x 3x (x (x 1 b. f(x = = x = x x Intercepts: f( =. f(x = when x =, 1. Vertical asymptote At x =, for the usual reason. Horizontal asymptote None. As x, f(x Divide to see non-horizontal asymptote: x 3x 6 = x 1 x x - 1 So we can see that y = x 1 is an oblique linear asymptote for y = f(x. (Since 6/(x - diminishes towards as x meanders out towards infinity. SIGNS: f(x is y = x 1 when x is between 1 (, esca csts - p
6 3 x 1 (x 1(x x 1 c. f(x = = x x x-intercept (1,** vertical asymptote at x = 3 x 1 1 f(x = = x x x y = x EPRP- p6... this shows the given f is x an amount (1, 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. And as x, the difference zero. x = This is asymptotic behavior (but not a linear asymptote. NO worries: The only asymptotes you will be asked to find are linear asymptotes!!! **of course, we determined that the remaining zeroes are complex, and thus not on the graph: x x 1 = 1± 1 = 1 ± 3 i x = x x 1 (x-(x3 Here we see x =, x = 3 make f(x = (x-int. d. f(x = = x (x (x Here we see x =, x = cause in the 1 1 f( = / = / is the y-intercept. f(x = x x 1 x domain, and, since the numerator is not at the same time, cause vertical asymptotes VAs at x = & x = Numerator is dominated by x... behaves like x when x Denominator is dominated by x... behaves like x when x So f(x should behave like x /x when x. (Horizontal asymptote at y=1. 1 /x 1/x f(x = 1 /x Another way to reach that conclusion... SIGNS: f(x is when x is between 3 One final investigation helps us determine an important feature affecting the graph s shape. Does the function cross its horizontal asymptote? Can x x 1 = 1 x We solve: x x 1 = x x 1 = x = x = / And see that f crosses its horizontal asymptote at (x =., 1 esca csts - p6
7 EPRP- p7 6. 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. 3 a. Find the value of P(x = x x x x 11 without raising 11 to a power. 11 This tells us that P(11 =, thus 11 is a zero, and (x 11 is a FACTOR of P. 3 3 In fact, this tells us: P(x = (x 11 ( x x x x x 1 3 b. If P(x = (x (x x 7x 13 7, what is P(? P( = ( ( whatever 7 = 7 SO: P( = 7... JUST as the remainder theorem states. (Go ahead, check it out! esca csts - p7
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