#### Transcript Chapter 2 Notes

HW: Pg. 175 #7-16 Polynomial Functions- ◦ Let n be a nonnegative integer and let a0, a1, a2,…, an-1, an be real numbers with an≠0. The functions given by f(x)=anxn + an-1xn-1+…+a2x2+a1x+a0 Is a polynomial function of degree n. The leading coefficient is an. f(x)=0 is a polynomial function. *it has no degree or leading coefficient. F(x) = 5x3-2x-3/4 G(x) = √(25x4+4x2) H(x) = 4x-5+6x K(x)=4x3+7x7 Name Form Degree Zero Function F(x) = 0 Undefined Constant Function F(x)=a (a≠0) 0 Linear Function F(x)=ax+b (a≠0) 1 Quadratic Function F(x)=ax2+bx+c (a≠0) 2 F(x) = ax+b Slope-Intercept form of a line: Find an equation for the linear function f such that f(-2) = 5 and f(3) = 6 The average rate of change of a function y=f(x) between x=a and x=b, a≠b, is [F(b)-F(a)]/[b-a] Weehawken High School bought a $50,000 building and for tax purposes are depreciating it $2000 per year over a 25-yr period using straight-line depreciation. 1. 2. 3. 4. What is the rate of change of the value of the building? Write an equation for the value v(t) of the building as a linear function of the time t since the building was placed in services. Evaluate v(0) and v(16) Solve v(t)=39,000 Point of View Characterization Verbal Polynomial of degree 1 Algebraic F(x)=mx+b (m≠0) Graphical Slant line with slope m and yintercept b Analytical Function with constant nonzero rate of change m, f is increasing if m>0, decreasing if m<0 Sketch how to transform f(x)=x2 into: G(x)=-(1/2)x2+3 H(x)=3(x+2) 2-1 If g(x) and h(x) and in the form f(x)=ax2+bx+c, what do you notice about g(x) and h(x) when a is a certain value (negative or positive)? f(x)=ax2+bx+c We want to find the axis of symmetry, which is x=-b/(2a). Then: The graph of f is a parabola with vertex (x,y), where x=-b/(2a). If a>0, the parabola opens upward, and if a<0, it opens downward. x=-b/(2a) F(x)=3x2+5x-4 G(x)=4x2+12x+4 H(x)=6x2+9x+3 f(x)=5x2+10x+5 Any Quadratic Function f(x)=ax2+bx+c, can be written in the vertex form: ◦ F(x)=a(x-h)2+k Where (h,k) is your vertex h=-b/(2a) and k=is the y F(x)=3x2+12x+11 =3(x2+4x)+11 =3(x2+4x+() - () )+11 =3(x2+4x+(2)2-(2)2)+11 =3(x2+4x+4)-3(4)+11 =3(x+2)2-1 f(x)=a(x-h)2+k Factor 3 from the x term Prepare to complete the square. Complete the square. Distribute the 3. F(x)=3x2+5x-4 F(x)=8x-x2+3 G(x)=5x2+4-6x Point of View Characterization Verbal Polynomial of degree ___ Algebraic F(x)=______________ (a≠0) Graphical a>0 a<0 HW: Pg.189 #1-10 F(x)=k*xa ◦ a is the power, k is the constant of variation EXAMPLES: Formulas Power Constant of Variation C=2∏r 1 2∏ A=∏r2 2 ∏ F(x)=4x3 G(x)=1/2x6 H(x)=6x-2 F(x) = ∛x 1/(x2) What type of Polynomials are these functions? (HINT: count the terms) 6cx-5 h/x4 4∏r2 3*2x ax 7x8/9 HW: Pg 203 #33-42e F(x)=x3+x G(x)=x3-x H(x)=x4-x2 Find local extrema and zeros for each polynomial F(x)=2x3 F(x)=-x3 F(x)=-2x4 F(x)=4x4 What do you notice about the limits of each function? F(x)=x3—2x2-15x What do these zeros tell us about our graph? F(x)=3x3 + 12x2 – 15x H(x)=x2 + 3x2 – 16 G(x)=9x3 - 3x2 – 2x K(x)=2x3 - 8x2 + 8x F(x)=6x2 + 18x – 24 HW: Pg. 216 #1-6 3587/32 (3x3+5x2+8x+7)/(3x+2) F(x) = d(x)*q(x)+r(x) F(x) and d(x) are polynomials where q(x) is the quotient and r(x) is the remainder (3x3+5x2+8x+7)/(3x+2) Write (2x4+3x3-2)/(2x2+x+1) in fraction form D(x)=x-k, degree is 1, so the remainder is a real number Divide f(x)=3x2+7x-20 by: ◦ (a) x+2 (b) x-3 (c) x+5 Remainder Theorem: If a polynomial f(x) is divided by x-k, then the remainder is r=f(k) Ex: (x2+3x+5)/(x-2) k=2 So, f(k)=f(2)=(2)2+3(2)+5=15=remainder Divide f(x)=3x2+7x-20 by: ◦ (a) x+2 (b) x-3 (c) x+5 If d(x)=x-k, where f(x)=(x-k)q(x) + r Then we can evaluate the polynomial f(x) at x=k: F(x)=2x2-3x+1; k=2 F(x)=2x3+3x2+4x-7; k=2 F(x)=x3-x2+2x-1; k=-3 Now we can use this method to find both remainders and quotients for division by x-k, called synthetic division. (2x3-3x2-5x-12)/(x-3) K becomes zero of divisor 3 | 2 -3 -5 -12 _____________ * Since the leading coefficient of the dividend must be the leading coefficient , copy the first “2” into the first quotient position. * Multiply the zero of the divisor (3) by the most recent coefficient of the quotient (2). Write the product above the line under the next term (-3). * Add the next coefficient of the dividend to the product just found and record sum below the line in the same column. * Repeat the “multiply” and “add” steps until the last row is completed. (x3-5x2+3x-2)/(x+1) (9x3+7x2-3x)/(x-10) (5x4-3x+1)/(4-x) Suppose f is a polynomial function of degree n1 of the form f(x)=anxn+…+a0 with every coefficient an integer. If x=p/q is a rational zero of f, where p and q have no common integer factors other than 1, then ◦ P is an integer factor of the constant coefficient a0, and ◦ Q is an integer factor of the leading coefficient an. Example: Find rational zeros of f(x)=x3-3x2+1 F(x)=3x3+4x2-5x-2 Potential Rational Zeros: F(x)=6x3-5x-1 F(x)=2x3-x2-9x+9 Let f be a polynomial function of degree n≥1 with a positive leading coefficient. Suppose f(x) is divided by x-k using synthetic division. If k≥0 and every number in the last line is nonnegative (positive or zero), then k is an upper bound for the real zeros of f. If k≤0 and the numbers in the last line are alternately nonnegative and nonpositive, then k is a lower bound for the real zeros of f. Lets establish that all the real zeros of f(x)=2x4-7x3-8x2+14x+8 must lie in the interval [-2,5] Establish bounds for real zeros Find the real zeros of a polynomial functions by using the rational zeros theorem to find potential rational zeros Use synthetic division to see which potential rational zeros are a real zero Complete the factoring of f(x) by using synthetic division again or factor. F(x)=10x5-3x2+x-6 F(x)=2x3-3x2-4x+6 F(x)=x3+x2-8x-6 F(x)=x4-3x3-6x2+6x+8 F(x)=2x4-7x3-2x2-7x-4 In the 17th century, mathematicians extended the definition of √(a) to include negative real numbers a. i =√(-1) is defined as a solution of (i )2 +1=0 For any negative real number √(a) = √|a|*i a +bi , where a, b are real numbers ◦ a+bi is in standard form Sum: (a+bi) +(c+di) = (a+c) + (b+d)i Difference: (a+bi) – (c+di) = (a-c) + (b-d)I EX: (a) (8 - 2i) + (5 + 4i) (b) (4 – i) – (5 + 2i) (2+4i)(5-i) Z=(1/2)+(√3/2)i, find Z2 Z = a+bi = a – bi When do we need to use conjugates? (2+3i)/(1-5i) ax2+bx+c=0 Try: x2-5x+11=0 Find all zeros: f(x) = x4 + x3 + x2 + 3x - 6 HW: Pg. 234-235 #2-10e, 28-34e Fundamental Theorem of Algebra – A polynomial function of degree n has n complex zeros (real and nonreal). Linear Factorization Theorem – If f(x) is a polynomial function of degree n>0, then f(x) has n linear factors and F(x) = a(x-z1)(x-z2)…(x-zn) Where a is the leading coefficient of f(x) and z1, z2, …, zn are the complex zeros of the function. X=k is a… K is a Factor of f(x): F(x)=(x-2i)(x+2i) F(x)=(x-3)(x-3)(x-i)(x+i) Suppose that f(x) is a polynomial function with real coefficients. If a+bi is a zero of f(x), then the complex conjugate a-bi is also a zero of f(x) 1. 2. 3. What can happen if the coefficients are not real? Use substitution to verify that x=2i and x=-i are zeros of f(x)=x2-ix+2. Are the conjugates of 2i and –i also zeros of f(x)? Use substitution to verify that x=i and x=1-i are zeros of g(x)=x2-x+(1+i). Are the conjugates of i and 1-i also zeros of g(x)? What conclusions can you draw from parts 1 and 2? Do your results contradict the theorem about complex conjugates? Given that -3, 4, and 2-i are zeros, find the polynomial: Given 1, 1+2i, 1-i, find the polynomial: The complex number z=1-2i is a zero of f(x)=4x4+17x2+14x+65, find the remaining zeros, and write it in its linear factorization. 3x5-2x4+6x3-4x2-24x+16 HW: Pg. 246 #19-30 F and g are polynomial functions with g(x)≠0. the functions: ◦ R(x)=f(x)/g(x) is a rational function ◦ Find the domain of : f(x)=1/(x+2) (a) g(x)=2/(x+3) (b) H(x)=(3x-7)/(x-2) Find horizontal and vertical asymptotes of f(x)=(x2+2)/(x2+1) Find asymptotes and intercepts of the function f(x)=x3/(x2-9) HW: Pg. 254 #7-17 X + 3/x = 4 2/(x-1) + x = 5 (2x)/(x-1) + 1/(x-3) = 2/(x2-4x+3) 3x/(x+2) + 2/(x-1) = 5/(x2+x-2) (x-3)/x + 3/(x+2) + 6/(x2 +2x) = 0 Find the dimensions of the rectangle with minimum perimeter if its area is 200 square meters. Find the least perimeter: A = 200 HW: Finish 2.9 WKSH F(x)=(x+3)(x2+1)(x-4)2 Determine the real number values of x that cause the function to be zero, positive, or negative: (x+3)(x2+1)(x-4)2 > 0 (x+3)(x2+1)(x-4)2 ≥ 0 (x+3)(x2+1)(x-4)2 < 0 (x+3)(x2+1)(x-4)2 ≤ 0 2x3-7x2-10x+24>0 Solve Graphically: x3-6x2≤2-8x *Plug function into your calculator* Section 2.9 #1-12 odd Check yourself