Jeff Bivin -- LZHS Graphs of Polynomial Functions By: Jeffrey Bivin Lake Zurich High School...
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Transcript of Jeff Bivin -- LZHS Graphs of Polynomial Functions By: Jeffrey Bivin Lake Zurich High School...
Jeff Bivin -- LZHS
Graphs of Polynomial Functions
By: Jeffrey Bivin
Lake Zurich High School
Last Updated: October 6, 2009
Jeff Bivin -- LZHS
Continuous FunctionThe graph of a polynomial function
f(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a0 is continuous if it has no breaks, holes,
vertical asymptotes or sharp turns.
Jeff Bivin -- LZHS
Leading Coefficient Test& End Behavior
f(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a0
n is even n is odd
If an is positive If an is positive
If an is negative If an is negative
Jeff Bivin -- LZHS
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 ↑ ↓
y y
y
y
y y
y
y
Jeff Bivin -- LZHS
Real Zeros & Relative Extrema
1. The function has at most n real zeros
2. The graph has at most n-1 relative extrema (relative minima or relative maxima)
f(x) = anxn + an-1xn-1 + an-2xn-2 + . . . + a1x + a0
Jeff Bivin -- LZHS
Finding Real Zeros
F(x) is a polynomial function….
1. x = a is a zero of the function
2. x = a is a solution of f(x) = 0
3. x – a is a factor of f(x)
4. (a, 0) is an x-intercept of the graph of f(x)
Jeff Bivin -- LZHS
Find all real zerosof the polynomial functions
1. F(x) = x2 – 121
2. F(x) = 2x2 – 14x + 24
3. F(x) =
4. F(x) = x4 – x3 – 20x2
5. F(x) = (x+2)4(x-3)5(x+4)3(x-5)2
25 163 3x + x + 4
Jeff Bivin -- LZHS
1. F(x) = x2 – 121 = 0
(x – 11)(x + 11) = 0
x – 11 = 0 or x + 11 = 0
x = 11 or x = – 11
Jeff Bivin -- LZHS
2. F(x) = 2x2 – 14x + 24 = 0
2(x – 4)(x – 3) = 0
2 = 0 or x – 4 = 0 or x – 3 = 0
x = 4 or x = 3
2(x2 – 7x + 12) = 0
Jeff Bivin -- LZHS
3. 25 163 3 4 0( )F x x x
213 5 16 12 0x x
13 5 6 2 0x x
13 0 5 6 0 2 0or x or x
65 2x or x
Jeff Bivin -- LZHS
4. F(x) = x4 – x3 – 20x2 = 0
x2(x – 5)(x + 4) = 0
x2 = 0 or x – 5 = 0 or x + 4 = 0
x = 0 or x = 5 or x = – 4
x2(x2 – x – 20) = 0
Jeff Bivin -- LZHS
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 = -2 or x = 3 or x = – 4 or x = 5
Jeff Bivin -- LZHS
Multiplicity – repeated zeros
• If the multiplicity is odd graph crosses x axis
• If the multiplicity is even graph touches x axis
Jeff Bivin -- LZHS
F(x) = (x-5)2(x-3)5(x-1)4(x+4)2(x+7)5
Degree = 2+5+4+2+5 = 18
(left side): as x - ∞, y + ∞(right side): as x + ∞, y + ∞
x = 5 multiplicity 2 touchx = 3 multiplicity 5 cross
x = 1 multiplicity 4 touchx = -4 multiplicity 2 touchx = -7 multiplicity 5 cross
TC T C T
Jeff Bivin -- LZHS
Find zeros of the polynomial function f(x) = x3 – 4x2 – 25x + 100
x3 – 4x2 – 25x + 100 = 0
x2(x – 4) – 25(x – 4) = 0
( x – 4)(x2 – 25) = 0
( x – 4)(x – 5)(x+5) = 0x – 4 = 0 x – 5 = 0 x + 5 = 0
x = 4 x = 5 x = –5
Jeff Bivin -- LZHS
Find a polynomial function that has the given zeros
Zeros: 5, 3, -2
x = 5 x = 3 x = –2x – 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 + 30 = 0
x3 – 6x2 – x + 30 = 0
f(x) = x3 – 6x2 – x + 30
Jeff Bivin -- LZHS
Find a polynomial function that has the given zeros
Zeros:
f(x) = x3 – 4x2 + 2x + 4 2 1 3 1 3, , 2 1 3 1 3 0x x x
2 1 3 1 3 0x x x
2 1 3 1 3 0x x x
222 1 1 3 1 3 3 0x x x x
22 2 1 3 0x x x
22 2 2 0x x x 3 2 22 2 2 4 4 0x x x x x
3 24 2 4 0x x x
Jeff Bivin -- LZHS
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).
a b
f(a)f(b)
Jeff Bivin -- LZHS
How is this especially helpful?
f(5) = 7
f(4) = -3
We must have a real zero
between the x values of 4 and
5
We may have more than one real zero in the interval [4, 5]