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HW: Pg. 175 #7-16

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Polynomial Functions- ◦ Let n be a nonnegative integer and let a 0, a 1, a 2,…, a n-1, a n be real numbers with a n ≠0. The functions given by f(x)=a n x n + a n-1 x n-1 +…+a 2 x 2 +a 1 x+a 0 Is a polynomial function of degree n. The leading coefficient is a n. f(x)=0 is a polynomial function. *it has no degree or leading coefficient.

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F(x) = 5x 3 -2x-3/4 G(x) = √(25x 4 +4x 2 ) H(x) = 4x -5 +6x K(x)=4x 3 +7x 7

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NameFormDegree Zero FunctionF(x) = 0Undefined Constant FunctionF(x)=a (a≠0)0 Linear FunctionF(x)=ax+b (a≠0)1 Quadratic FunctionF(x)=ax 2 +bx+c (a≠0)2

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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

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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]

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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. What is the rate of change of the value of the building? 2. 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. 3. Evaluate v(0) and v(16) 4. Solve v(t)=39,000

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Point of ViewCharacterization VerbalPolynomial of degree 1 AlgebraicF(x)=mx+b (m≠0) GraphicalSlant line with slope m and y- intercept b AnalyticalFunction with constant nonzero rate of change m, f is increasing if m>0, decreasing if m<0

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Sketch how to transform f(x)=x 2 into: G(x)=-(1/2)x 2 +3 H(x)=3(x+2) 2 -1 If g(x) and h(x) and in the form f(x)=ax 2 +bx+c, what do you notice about g(x) and h(x) when a is a certain value (negative or positive)?

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f(x)=ax 2 +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.

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x=-b/(2a)

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F(x)=3x 2 +5x-4 G(x)=4x 2 +12x+4 H(x)=6x 2 +9x+3 f(x)=5x 2 +10x+5

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Any Quadratic Function f(x)=ax 2 +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

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F(x)=3x 2 +12x+11 f(x)=a(x-h) 2 +k =3(x 2 +4x)+11 Factor 3 from the x term =3(x 2 +4x+() - () )+11 Prepare to complete the square. =3(x 2 +4x+(2) 2 -(2) 2 )+11 Complete the square. =3(x 2 +4x+4)-3(4)+11 Distribute the 3. =3(x+2) 2 -1

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F(x)=3x 2 +5x-4 F(x)=8x-x 2 +3 G(x)=5x x

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Point of ViewCharacterization VerbalPolynomial of degree ___ AlgebraicF(x)=______________ (a≠0) Graphicala>0 a<0

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HW: Pg.189 #1-10

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F(x)=k*x a ◦ a is the power, k is the constant of variation EXAMPLES: FormulasPowerConstant of Variation C=2∏r12∏ A=∏r 2 2∏ F(x)=4x 3 G(x)=1/2x 6 H(x)=6x -2

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F(x) = ∛x 1/(x 2 ) What type of Polynomials are these functions? (HINT: count the terms)

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6cx -5 h/x 4 4∏r 2 3*2 x ax 7x 8/9

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HW: Pg 203 #33-42e

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F(x)=x 3 +x G(x)=x 3 -x H(x)=x 4 -x 2 Find local extrema and zeros for each polynomial

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F(x)=2x 3 F(x)=-x 3 F(x)=-2x 4 F(x)=4x 4 What do you notice about the limits of each function?

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F(x)=x 3 —2x 2 -15x What do these zeros tell us about our graph?

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F(x)=3x x 2 – 15x H(x)=x 2 + 3x 2 – 16 G(x)=9x 3 - 3x 2 – 2x K(x)=2x 3 - 8x 2 + 8x F(x)=6x x – 24

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HW: Pg. 216 #1-6

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3587/32 (3x 3 +5x 2 +8x+7)/(3x+2)

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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

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(3x 3 +5x 2 +8x+7)/(3x+2) Write (2x 4 +3x 3 -2)/(2x 2 +x+1) in fraction form

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D(x)=x-k, degree is 1, so the remainder is a real number Divide f(x)=3x 2 +7x-20 by: ◦ (a) x+2 (b) x-3 (c) x+5

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Remainder Theorem: If a polynomial f(x) is divided by x-k, then the remainder is r=f(k) Ex: (x 2 +3x+5)/(x-2) k=2 So, f(k)=f(2)=(2) 2 +3(2)+5=15=remainder

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Divide f(x)=3x 2 +7x-20 by: ◦ (a) x+2 (b) x-3 (c) x+5

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If d(x)=x-k, where f(x)=(x-k)q(x) + r Then we can evaluate the polynomial f(x) at x=k:

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F(x)=2x 2 -3x+1; k=2 F(x)=2x 3 +3x 2 +4x-7; k=2 F(x)=x 3 -x 2 +2x-1; k=-3

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Now we can use this method to find both remainders and quotients for division by x-k, called synthetic division. (2x 3 -3x 2 -5x-12)/(x-3) K becomes zero of divisor

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3 | _____________ * 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.

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(x 3 -5x 2 +3x-2)/(x+1) (9x 3 +7x 2 -3x)/(x-10) (5x 4 -3x+1)/(4-x)

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Suppose f is a polynomial function of degree n1 of the form f(x)=a n x n +…+a 0 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 a 0, and ◦ Q is an integer factor of the leading coefficient a n. Example: Find rational zeros of f(x)=x 3 -3x 2 +1

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F(x)=3x 3 +4x 2 -5x-2 Potential Rational Zeros:

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F(x)=6x 3 -5x-1 F(x)=2x 3 -x 2 -9x+9

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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.

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Lets establish that all the real zeros of f(x)=2x 4 -7x 3 -8x 2 +14x+8 must lie in the interval [-2,5]

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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.

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F(x)=10x 5 -3x 2 +x-6

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F(x)=2x 3 -3x 2 -4x+6 F(x)=x 3 +x 2 -8x-6 F(x)=x 4 -3x 3 -6x 2 +6x+8 F(x)=2x 4 -7x 3 -2x 2 -7x-4

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In the 17 th 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

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a +bi, where a, b are real numbers ◦ a+bi is in standard form

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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)

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(2+4i)(5-i) Z=(1/2)+(√3/2)i, find Z 2

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Z = a+bi = a – bi When do we need to use conjugates?

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(2+3i)/(1-5i)

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ax 2 +bx+c=0

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Try: x 2 -5x+11=0

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Find all zeros: f(x) = x 4 + x 3 + x 2 + 3x - 6

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HW: Pg #2-10e, 28-34e

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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-z 1 )(x-z 2 )…(x-z n ) Where a is the leading coefficient of f(x) and z 1, z 2, …, z n are the complex zeros of the function.

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X=k is a… K is a Factor of f(x):

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F(x)=(x-2i)(x+2i) F(x)=(x-3)(x-3)(x-i)(x+i)

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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)

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What can happen if the coefficients are not real? 1. Use substitution to verify that x=2i and x=-i are zeros of f(x)=x 2 -ix+2. Are the conjugates of 2i and –i also zeros of f(x)? 2. Use substitution to verify that x=i and x=1-i are zeros of g(x)=x 2 -x+(1+i). Are the conjugates of i and 1-i also zeros of g(x)? 3. What conclusions can you draw from parts 1 and 2? Do your results contradict the theorem about complex conjugates?

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Given that -3, 4, and 2-i are zeros, find the polynomial: Given 1, 1+2i, 1-i, find the polynomial:

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The complex number z=1-2i is a zero of f(x)=4x 4 +17x 2 +14x+65, find the remaining zeros, and write it in its linear factorization.

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3x 5 -2x 4 +6x 3 -4x 2 -24x+16

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HW: Pg. 246 #19-30

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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)

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(a) g(x)=2/(x+3) (b) H(x)=(3x-7)/(x-2)

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Find horizontal and vertical asymptotes of f(x)=(x 2 +2)/(x 2 +1) Find asymptotes and intercepts of the function f(x)=x 3 /(x 2 -9)

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HW: Pg. 254 #7-17

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X + 3/x = 4 2/(x-1) + x = 5

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(2x)/(x-1) + 1/(x-3) = 2/(x 2 -4x+3) 3x/(x+2) + 2/(x-1) = 5/(x 2 +x-2)

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(x-3)/x + 3/(x+2) + 6/(x 2 +2x) = 0

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Find the dimensions of the rectangle with minimum perimeter if its area is 200 square meters. Find the least perimeter: A = 200

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HW: Finish 2.9 WKSH

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F(x)=(x+3)(x 2 +1)(x-4) 2 Determine the real number values of x that cause the function to be zero, positive, or negative:

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(x+3)(x 2 +1)(x-4) 2 > 0 (x+3)(x 2 +1)(x-4) 2 ≥ 0 (x+3)(x 2 +1)(x-4) 2 < 0 (x+3)(x 2 +1)(x-4) 2 ≤ 0

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2x 3 -7x 2 -10x+24>0 Solve Graphically: x 3 -6x 2 ≤2-8x *Plug function into your calculator*

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Section 2.9 #1-12 odd Check yourself

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