**Unformatted text preview: **Complete the Square -x2 + 8x + 10 = 0 -(x +4) = 2 2 -x + 8x + __ = -10 -x = -2 -x2 + 8x + 16 = (-10) + (16) -(x + 4)2 = 4 Inequalities -open interval: (a, b); a < x < b -a < x (a, ) -closed interval: [a, b]; a x b -a x [a, ) -half-open interval: (a, b] or [a, b) -b > x (-, b) a < x b or a x < b -b x (-, b] Graphs -vertex is at [-b/2a, f(-b/2a)] -axis of symmetry is x = (-b/2a) -x-intercept is f(x) = 0 -y-intercept is f(x) = c Linear Functions -C(x) = cx +F; F = fixed costs (cost function) -R(x) = sx (revenue function) -P(x) = R(x) C(x) (profit function) -break-even = P(x) = 0, or R(x) = C(x) (where R(x) and C(x) intersect) Compound Interest -A = P[1 + (r/n)]nt -A = Pert (continually compounded interest) Logarithms -logb(mn) = logb(m) + logb(n) -log3(81) = 4, 34 = 81 -logb(m/n) = logb(m) logb(n)-log4(16) = 2, 42 = 16 -logb(mn) = (n)logb(m) -log2(16) = 4, 24 = 16 -logb(1) = 0 -logb(b) = 1 Exponential Functions as Models -F(t) = F0ekt; F0 and k are constants -F(0) = F0 -F(t) = F0e-kt (exponential decay) -for half-life, = F0e-kt; t = half-life -L(t)= C Ae-kt (learning curve) -F(t) = A / 1 + Be-kt (logistic growth) Annuities -(future value) S = R (1 + (r/n))n -1 (r/n) -(present value) P = R 1- (1 + (r/n))n (r/n) Linear Programming Problem Example -A company manufactures two products (A & B) and two machines (I & II). The company will make a profit of $3/unit on product A and $4/unit on product B. To manufacture product A, machine I requires 6 minutes/unit and machine II requires 5 minutes/unit. To manufacture product B, machine I requires 9 minutes/unit and machine II requires 4 minutes/unit. There is 5 hours of machine time available on machine I and 3 hours on machine II. How many units of each should be produced to maximize profits? -Set Up: -P(x) = 3A + 4B (maximize) -machine I time = 6A + 9B 300 -machine II time = 5A + 4B 180 ...

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