7.1 Polynomial Functions Degree and Lead Coefficient End Behavior.
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Transcript of 7.1 Polynomial Functions Degree and Lead Coefficient End Behavior.
Polynomial should be written in descending order
The polynomial is not in the correct order3x3 + 2 – x5 + 7x2 + x
Just move the terms around
-x5 + 3x3 + 7x2 + x + 2
Now it is in correct form
When the polynomial is in the correct order
Finding the lead coefficient is the number in front of the first term
-x5 + 3x3 + 7x2 + x + 2
Lead coefficient is – 1It degree is the highest degree
Degree 5Since it only has one variable, it is a
Polynomial in One Variable
Evaluate a Polynomial
To Evaluate replace the variable with a given value.
f(x) = 3x2 – 3x + 1 Let x = 4, 5, and 6
Evaluate a Polynomial
To Evaluate replace the variable with a given value. f(x) = 3x2 – 3x + 1 Let x = 4, 5, and 6
f(4) = 3(4)2 – 3(4) + 1 = 37 = 3(16) – 12 + 1 = 48 – 12 + 1 = 36 + 1 = 37
Evaluate a Polynomial
To Evaluate replace the variable with a given value.
f(x) = 3x2 – 3x + 1 Let x = 4, 5, and 6
f(4) = 3(4)2 – 3(4) + 1 = 37
f(5) = 3(5)2 – 3(5) + 1 =
Evaluate a Polynomial
To Evaluate replace the variable with a given value. f(x) = 3x2 – 3x + 1 Let x = 4, 5, and 6
f(4) = 3(4)2 – 3(4) + 1 = 37f(5) = 3(5)2 – 3(5) + 1 = 61
= 3(25) – 15 + 1 = 75 – 15 + 1 = 61
Evaluate a Polynomial
To Evaluate replace the variable with a given value.
f(x) = 3x2 – 3x + 1 Let x = 4, 5, and 6
f(4) = 3(4)2 – 3(4) + 1 = 37
f(5) = 3(5)2 – 3(5) + 1 = 61
f(6) = 3(6)2 – 3(6) + 1 =
Evaluate a Polynomial
To Evaluate replace the variable with a given value. f(x) = 3x2 – 3x + 1 Let x = 4, 5, and 6
f(4) = 3(4)2 – 3(4) + 1 = 37f(5) = 3(5)2 – 3(5) + 1 = 61f(6) = 3(6)2 – 3(6) + 1 = 91
= 3(36) – 18 + 1 = 91
Find b(2x – 1) – 3b(x) if b(m) = 2m2 + m - 1
Do this problem in two parts
b(2x – 1) = 2(2x – 1)2 + (2x -1) – 1
=2(2x – 1)(2x – 1) + (2x – 1) – 1
=2(4x2 – 2x -2x + 1) + (2x -1) – 1
= 2(4x2 – 4x + 1) + (2x – 1) -1
= 8x2 – 8x + 2 + 2x -1 – 1
= 8x2 - 6x
Find b(2x – 1) – 3b(x) if b(m) = 2m2 + m - 1
Do this problem in two parts
b(2x – 1) = 8x2 - 6x
-3b(x) = -3(2x2 + x – 1) = -6x2 – 3x + 3
b(2x – 1) – 3b(x) = (8x2 – 6x) + (-6x2 – 3x + 3)
= 2x2 – 9x + 3
End Behavior We understand the end behavior of a quadratic
equation.
y = ax2 + bx + c both sides go up if a> 0
both sides go down a < 0
If the degree is an even number it will always be the same. y = 6x8 – 5x3 + 2x – 5
go up since 6>0 and 8 the degree is even
End Behavior If the degree is an odd number it will
always be in different directions. y = 6x7 – 5x3 + 2x – 5
Since 6>0 and 7 the degree is odd
raises up as x goes to positive infinite
and falls down as x goes to negative infinite.
End Behavior If the degree is an odd number it will always
be in different directions.
y = -6x7 – 5x3 + 2x – 5
Since -6<0 and 7 the degree is odd
falls down as x goes to positive infinite
and raises up as x goes to negative infinite.
End Behavior
If a is positive and degree is even, then the polynomial raises up on both ends (smiles)
If a is negative and degree is even, then the polynomial falls on both ends (frowns)
End Behavior
If a is positive and degree is odd, then the polynomial raises up as x becomes larger, and falls as x becomes smaller
If a is negative and degree is odd, then the polynomial falls as x becomes larger, and rasies as x becomes smaller
Tell me if a is positive or negative and if the degree is
even or odda is positive and the degree is odd
Tell me if a is positive or negative and if the degree is
even or odda is positive and the degree is even
Tell me if a is positive or negative and if the degree is
even or odda is negative and the degree is odd