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Graphs of Polynomial Functions

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Presentation on theme: "Graphs of Polynomial Functions"— Presentation transcript:

1 Graphs of Polynomial Functions
By: Jeffrey Bivin Lake Zurich High School Last Updated: October 6, 2009

2 Continuous Function The graph of a polynomial function f(x) = anxn + an-1xn-1 + an-2xn a1x + a0 is continuous if it has no breaks, holes, vertical asymptotes or sharp turns.

3 Are these graphs continuous?
NO NO NO NO YES YES However, this is not a polynomial function

4 Leading Coefficient Test & End Behavior
f(x) = anxn + an-1xn-1 + an-2xn a1x + a0 n is even n is odd If an is positive If an is positive If an is negative If an is negative

5 Leading Coefficient Test & End Behavior
left (x  ) right (x  ) f(x) = 5x4 + 7x2 + 3 f(x) = -7x8 + x + 3 f(x) = 5x5 + 7x2 + 3x f(x) = -3x5 + 6x3 + 2

6 Real Zeros & Relative Extrema
f(x) = anxn + an-1xn-1 + an-2xn a1x + a0 1. The function has at most n real zeros 2. The graph has at most n-1 relative extrema (relative minima or relative maxima)

7 Finding Real Zeros Terminology F(x) is a polynomial function….
x = a is a zero of the function x = a is a solution of f(x) = 0 x – a is a factor of f(x) (a, 0) is an x-intercept of the graph of f(x)

8 Find all real zeros of the polynomial functions
F(x) = x2 – 121 F(x) = 2x2 – 14x + 24 F(x) = F(x) = x4 – x3 – 20x2 F(x) = (x+2)4(x-3)5(x+4)3(x-5)2

9 1. F(x) = x2 – 121 = 0 (x – 11)(x + 11) = 0 x – 11 = 0 or x + 11 = 0

10 2. F(x) = 2x2 – 14x + 24 = 0 X 2(x2 – 7x + 12) = 0 2(x – 4)(x – 3) = 0
2 = 0 or x – 4 = 0 or x – 3 = 0 x = or x = 3

11 3. X

12 4. F(x) = x4 – x3 – 20x2 = 0 x2(x2 – x – 20) = 0 x2(x – 5)(x + 4) = 0
x2 = 0 or x – 5 = 0 or x + 4 = 0 x = or x = or x = – 4

13 5. F(x) = (x+2)4(x-3)5(x+4)3(x-5)2 = 0
x + 2 = 0 or x – 3 = 0 or x + 4 = 0 or x – 5 = 0 x = or x = 3 or x = – 4 or x = 5

14 Graph F(x) = (x-2)2 F(x) = (x-2)3 F(x) = (x-2)4 F(x) = (x-2)5

15 Multiplicity – repeated zeros
If the multiplicity is odd  graph crosses x axis If the multiplicity is even  graph touches x axis

16 F(x) = (x-5)2(x-3)5(x-1)4(x+4)2(x+7)5
Degree = = 18 (left side): as x  - ∞, y  + ∞ (right side): as x  + ∞, y  + ∞ x = 5 multiplicity 2  touch x = 3 multiplicity 5  cross x = 1 multiplicity 4  touch x = -4 multiplicity 2  touch C T T C T x = -7 multiplicity 5  cross

17 Find zeros of the polynomial function f(x) = x3 – 4x2 – 25x + 100

18 Find a polynomial function that has the given zeros
f(x) = x3 – 6x2 – x + 30 x = 5 x = 3 x = –2 x – 5 = 0 x – 3 = 0 x + 2 = 0 ( x – 5)(x – 3)(x + 2) = 0 (x – 5)(x2 – x – 6) = 0 x3 – x2 – 6x – 5x2 + 5x = 0 x3 – 6x2 – x = 0

19 Find a polynomial function that has the given zeros
f(x) = x3 – 4x2 + 2x + 4

20 Intermediate Value Theorem
Let a and b be real numbers such that a<b. If f(x) is a polynomial function such that f(a) ≠ f(b), then in the interval [a, b], f(x) takes on every value between f(a) and f(b). What does it mean? f(b) f(a) a b

21 How is this especially helpful?
We must have a real zero between the x values of 4 and 5 f(5) = 7 We may have more than one real zero in the interval [4, 5] f(4) = -3

22 Another Look……. A Table Perspective x y1 -2 9 -1 -5 1 3 2 12 7 4 -6
-5 1 3 2 12 7 4 -6 At least one real zero between & -1 sign change At least one real zero between & 1 sign change At least one real zero between & 4 sign change


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