Polynomial FunctionsChapter 2 Part 1

Standard Formf(x)=ax2+bx+c

Vertex Formf(x)=a(x-h)2+k

Intercept Formf(x)=a(x-d)(x-e)

y-int (0, c) let x = 0

then solve for f(x)let x = 0

then solve for f(x)

x-int(s)

let f(x) = 0 then factor or use quadratic

formula

let f(x) = 0 then solve for x (d, 0) & (e, 0)

Vertex

yv = substitute xv

and solve for y

(h, k)yv = substitute

xv and solve for

y

Quadratic Functions

Quadratic FormulaParabola opens up if a > 0 (pos); opens down if a < 0 (neg)

Parabola is symmetrical about the vertical line passing through the vertex

2

edxv

a

acbbx

2

42

a

bxv 2

Features of Polynomial Graphs

Polynomial graphs are continuousDegree of the polynomial (n)

◦n = the largest exponentEnd Behaviory-intercept:

◦ let x = 0 then evaluate for yZero(s) (aka x-intercepts) of the function:

◦ let y or f(x) = 0 then solve for xTurning Points (aka Relative Extrema):

◦estimate (algebraic method will be learned in Calculus)

Positive Lead Coefficient Negative Lead Coefficient

Odd Exponent

Think: y = x

Down on left; Up on right

f(x) ⇒ - ∞ as x ⇒ - ∞f(x) ⇒ + ∞ as x ⇒ + ∞

Think: y = - x

Up on left; Down on right

f(x) ⇒ + ∞ as x ⇒ - ∞f(x) ⇒ - ∞ as x ⇒ + ∞

Even Exponent

Think: y = x2

Up on both sides

f(x) ⇒ + ∞ as x ⇒ - ∞f(x) ⇒ + ∞ as x ⇒ + ∞

Think: y = - x2

Down on both sides

f(x) ⇒ - ∞ as x ⇒ - ∞f(x) ⇒ - ∞ as x ⇒ + ∞

Leading Term Test (aka End Behavior)

Zeros and Turning PointsDegree of the polynomial (n)

◦n = the largest exponent

Maximum # of zeros = n◦ zeros with odd multiplicity pass through x-

axis◦ zeros with even multiplicity turn at x-axis

Maximum # of turning points = n-1

Long Division of Polynomials

8657÷21

(8x3+6x2+5x+7)÷(2x+1)

1. Divide2. Multiply3. Subtract4. Repeat5. + Remainder

Divisor

1. Divide first term inside by first term outside

2. Multiply answer by divisor3. Subtract by changing to “add

opposite”4. Repeat until degree inside <

degree outside5. + Remainder

Divisor

Synthetic Division of Polynomials(a4x4 + a3x3 + a2x2 + a1x + a0) ÷ (x –

k) coefficients of dividendk

b3 b2 b1 b0 r

coeff of quotientdegree of quotient is one less than degree of

dividend

Interpreting the results:

a4 a3 a2 a1 a0

remainder

Vertical: AddDiagonal: multiply by k

Remainder and Factor TheoremsRemainder Theorem

◦If f(x) is divided by (x-k) then f(k) = r

Factor Theorem◦(x-k) is a factor of f(x) if and only if

f(k) = 0

Zeros of Polynomial Functions

If a is a zero of f(x) then:x=a is a solution of f(x) = 0(x – a) is a factor of f(x)(a, 0) is an x-intercept on the

graph of f(x)◦if a is a real number

Rational Zero TestIf f(x) is a polynomial function

f(x) = anxn + an-1xn-1 + … + a1x + a0

with integer coefficients, then all rational zeros of f(x) will have the form:

p = ± factors of constant term (a0) q ± factors of leading coefficient (an)

Fundamental Theorem of Algebra

A polynomial of degree n has exactly n complex zeros (including repeated zeros).

Complex Conjugate TheoremComplex zeros of a polynomial

function occur in conjugate pairs. In other words: if (a + bi) is a zero then (a – bi) is

also a zero

Upper and Lower Bound Tests

Upper Bound Test ◦ Works only for positive values of k◦ Indicates where the graph has reached its “end

behavior” on the right (no need to test larger k values)

◦ Signs of Quotient and Remainder are all pos or all neg

Lower Bound Test◦ Works only for negative values of k◦ Indicates where the graph has reached its “end

behavior” on the left (no need to test larger negative k values)

◦ Signs of Quotient and Remainder alternate between pos and neg

Zero can be thought of as either pos or neg

Descartes’ Rule of SignsIf f(x) is a polynomial function

with real coefficients, then◦the number of positive real zeros is

equal to the number of sign changes in f(x) or less than that by an even number.

◦the number of negative real zeros is equal to the number of sign changes in f(-x)

◦ or less than that by an even number.