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Review: Polynomial Functions (1st topic covered; Chapter 2 in text) 1. For each of the following functions, state the degree, and all roots in simplest radical form if possible. Then make a rough sketch. Location of local max/min. points are not necessary. a) f ( x) ( x 4)2 ( x 2 6 x 9) b) f ( x) ( x 2 9)( x 2 4 x 3) c) f ( x) x3 4 x 2 x 4 a) quartic function (4th degree). f(x) = - (x+4)(x+4)(x+3)(x+3) So the roots are x = -4, -4, -3, -3 Sketch: b) quartic function f(x) = (x+3)(x-3)(x+3)(x+1) So the roots are x = -3, -3, -3, -1 Sketch: c) cubic function f(x) = (x-1)(x-4)(x+1) (Use synthetic division to get these factors) So the roots are x = 1, 4, -1 Sketch: 2. Solve using a sign chart. Use interval notation. x 3 7 x 2 4 x 28 0 What does answer represent, graphically? Find the roots and do a sign chart. Using synthetic division, the factors are: f(x) = (x-2)(x2 – 5x – 14) = (x-2)(x-7)(x+2) So the roots are x = 2, 7, -2 The sign chart is - -2 + 2 7 + The final answer is x ε (-2, 2] U [7, ∞) It represents where the graph of y = x3 – 7x2 – 4x + 28 is above the x –axis. 3. List all possible rational roots using the rational root theorem. Then factor to find all roots. a) g ( x) x 4 7 x 2 12 b) h( x) x3 x 2 16 x 20 a) Possible rational roots are ±1, 2, 3, 4, 6, 12 Roots are x = 2, 2, -2, -2 (Use synthetic division or “quadratic substitution”… ie. let x2 = a and solve that equation). b) Possible rational roots are ±1,2,4,5,10,20 Roots are x = -5, 2, 2 (Use synthetic division) 4. A cubic function goes through the following points: Find the equation of the cubic function that fits the data. You may x y use substitution, elimination, or inverse matrices. 0 0 1 -12 Since y = ax3 + bx2 + cx + d, we have: 3 6 6 48 (0, 0) gives 0 = (0)a + (0)b + (0)c + d or in other words, d = 0 (1, -12) gives –12 = (1)a + (1)b + (1)c + d or –12 = a + b + c (3, 6) gives 6 = (3)3a + (3)2b + 3c + d or 6 = 27a + 9b + 3c (6, 48) gives 48 = (6)3a + (6)2b + (6)c + d or 48 = 216a + 36b + 6c Using one of the three methods listed above, we find that a = -1, b = 11, c = -22 and we know already that d = 0. So the final equation is: y = -x3 + 11x2 – 22x 5. If f(x) = 3x + 5, g(x) = x2 and h(x) = 2x, find a) f(g(2)) b) h(g(x)) c) f(g(h(1))) a) f ( g (2)) 3((2) 2 ) 5 17 b) h( g ( x)) 2 x 2 c) f ( g (h(1))) 3 2(1) 5 17 2 6. Find the equation of the tangent to y x3 4 x 10 at x = -1. Remember to express the answer in the form y = mx+b. The slope of the tangent is y’ = 3x2 + 4 using the power rule. At x = -1, this equals 7. That represents a positive and pretty steep slope! When x = -1, y = (-1)3 + 4(-1) + 10 = 5 I used the ORIGINAL function to find this y value, because I am finding a point on the curve. Its ordered pair is (-1, 5) So in y = mx + b, we know that m = 7, x = -1 and y = 5. Find b! 5 = 7(-1) + b b = 12 So the answer is y = 7x + 12 Graphically, here’s what the equation of the tangent line looks like against the original function: 7. Given f ( x) x3 12 x , sketch an accurate graph showing the roots (exact), location of local max./min. points (you will have to find f’(x) and set it equal to 0 to do this) Then state the domain, range, intervals of increase and decrease and intervals where the function is above and below the x-axis. To get the roots, factor. f(x) = x(x2 – 12) 0 = x(x2 – 12) x = 0, x = ±√12 (this is x = ±2√3 in simplest radical form, approx. ±3.46) To get the local max./min. points, find f’(x). The slope will be 0 at the max/min, so set f’(x) = 0 and solve for x. f’(x) = 3x2 – 12 0 = 3x2 – 12 0 = 3(x2 – 4) x = ±2 When x = 2, y = -16 in the original function. When x = -2, y = 16 in the original function. So the local max. is at (-2, 16) and the local min. is at (2, -16) Sketch: Analysis: D: x ε R R: y ε R Inc: x ε (-∞, -2) U (2, ∞) Dec: x ε (-2, 2) Above x-axis (f(x) > 0): x ε (-2√3, 0) U (2√3, ∞) Below x-axis (f(x) < 0): x ε (-∞, -2√3) U (0, 2√3) 8. Given that y x 2 3x +5, a) Find y’ using the power rule. b) Find y’ using the definition of the derivative, ie. limits: f ( x h) f ( x ) y ' lim h 0 h c) State the equation of the tangent line to this function at (3, 20) a) y’ = 2x + 3 b) ( x h) 2 3( x h) 5 ( x 2 3 x 5) h 2 2 x 2hx h 3 x 3h 5 x 2 3x 5 lim h0 h 2 2hx h 3h lim h0 h h(2 x h 3) lim h0 h lim h0 2 x h 3 y ' lim h0 2x 3 c) y = mx + b where x = 3, y = 20 and m = 9 (m = 2x + 3 where x = 3) So b must equal –7. The final answer is y = 9x – 7 1 x 1, , 5 and the function 3 has a y-intercept of . State the equation of this function in general form. 9. Suppose the roots of a quartic function are In factored form, the equation of this function would be f ( x) k ( x 1)(3x 1)( x 2 5) Notice how I changed all the given roots into their corresponding factors such that f(x) = 0 to find the roots. Note the “k” out front… constants factored out don’t affect the roots. But they can affect the y-intercept. Our y-intercept is 25. Substituting x = 0 to obtain the y-intercept, we have k 0 1 3(0) 1 0 2 5 25 k (1)(1)(5) 25 5k 25 k 5 Now expand the factored form to get it in general form.