Precalculus 2.6 Rational Functions and Asymptotes

2.6 Rational Functions and Asymptotes

A rational function can be written in the form:

f(x) = N(x)/D(x)

where N(x) is the numerator and
D(x) is the denominator and both are polynomials but D(x) is not the zero polynomial (because you cannot divide by zero)

I. Asymptotes of a Rational Function:
Let "f" be the rational function f(x) where

f(x) = N(x)/D(x) where

N(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀ and

D(x) = b_mx^m + b_m-1x^m-1 + ... + b₁x + b₀

Where N(x) and D(x) have no common factors.
1. The graph of "f" has vertical asymptotes at the zeros of D(x).
2. The graph of "f" has at most one horizontal asymptote determined by comparing the degrees of N(x) and D(x).
a. If n is less than m, the line y = 0 (x-axis) is a horizontal asymptote.
b. If n = m, the line y = (a_n)/(b_m) is a horizontal asymptote.
c. If n is greater than m, the graph of "f" has no horizontal asymptote.

Examples:
a) f(x) = (5x² + 3)/(-6x³ + 2x + 4)

n = 2 and m = 3 which means that n is less than m so the horizontal asymptote is y = 0

b) f(x) = (2x³ + 2x²)/(3x³ + 4x)

n = 3 and m = 3 so the line y = 2/3 is a horizontal asymptote.

c) f (x) = (3x⁴ + 2x² + 5)/(4x³ + 3x)

n = 4 and m = 3 so n is less than m, therefore there is not a horizontal asymptote.

II. Find the Domain and Asymptotes of f(x) = 3/ ((x - 2)³)

1. Find the vertical asymptotes by taking D(x) and setting it equal to zero.

(x - 2)³ = 0
x - 2 = 0
x = 2

therefore a vertical asymptote is the line x = 2

2. Find the horizontal asymptotes
n = 0 and m = 3 so n is less than m, therefore the line y = 0 is the horizontal asymptote.

3. Graph it using a graphing utility.

4. Domain (- ∞, 2) U (2, ∞)

5. Range (- ∞, 0) U (0, ∞)

6. check using a table:

{ (-3, -0.24), (-2, -.047), (-1, -.11), (0, -3.75), (1, -3), (2, error), (3, 3)}

Homework: Quiz (19, 20)
Pg. 187/ 7, 21, 23, 43*Pg. 195/ 13 – 19 odd, 27 – 31 odd, 35

Precalculus 2.5 The Fundamental Theorem of Algebra

2.5 The Fundamental Theorem of Algebra –
Proved by Carl Friedrich Gauss

If f (x) is a polynomial of a degree “n”, where n is greater than 0,
“ f ” has at least one zero in the complex number system.

I. Linear Factorization Theorem:
If f(x) is a polynomial of degree “n” where “n is greater than zero”, “f” has precisely “n” linear factors.

f (x) = a_n (x – c₁)( x – c₂)( x – c₃) … ( x – c_n) where
c₁, c₂, c₃, … c_n are complex numbers.

Factors of a Polynomial:

Example 1: f (x) = t⁴ - 5t³ + 15t² - 45t + 54

4th degree n = 4, n is greater than zero is true

Step 1: find all the possible zeros of the function:
(factors of 54)/(factors of 1) = ±1, ± 2, ± 3, ±6, ± 9, ± 18, ± 27, ± 54

Step 2: Use synthetic division to find out which ones are factors


f (x) = (x – 2)(x – 3)(x² + 9)
0 = (x – 2)(x – 3)(x² + 9)

0 = x – 2
x = 2

0 = x – 3
x = 3

0 = x² + 9
- 9 = x²
±√(-9) = x
± 3i = x

So to write this function in linear form:

f(x) = (x – 2)(x – 3)(x + 3i)(x – 3i)

So the following zeros of “f” are:

x = {2, 3, 3i, and -3i}

II. Complex Zeros Occur in Conjugate Pairs

In the last example, “± 3i”, the pair is 3i and -3i so we can conclude:

Let f(x) be a polynomial function that has real coefficients.
If a + bi, where b ≠ 0, is a zero of the function,
the conjugate a – bi is also a zero of the function.

Example: Your zeros are {2, 4 + i, and 4 – i}, find the polynomial function:

f(x) = (x – 2)(x – (4 + i))(x – (4 – i))
= (x – 2)((x – 4) – i)((x – 4) + i)
= (x – 2)( (x – 4)² - (i)²)
= (x – 2)(x² - 8x + 16 – (-1))
= (x – 2)(x² - 8x + 17)
= x³ – 8x² + 17x – 2x² + 16x – 34
= x³ – 10x² + 33x – 34

III. If you are given one zero, can you find the rest?

g(x) = 4x³ + 23x² + 34x – 10 given zero: -3 + i

Recall both conjugates:
– 3 + i and – 3 – i so


(x + 3 – i) (x + 3 + i)
= ((x + 3) – i)((x + 3) + i)
= ((x + 3)² - (i)²)
= x² + 6x + 9 – (-1)
= x² + 6x + 10

So using long division:


Every polynomial of degree n is greater than zero with real coefficients can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros.

A quadratic factor with no real zeros is said to be irreducible over the real numbers.

x² + 1 = (x + i)(x – i) is irreducible over the real numbers
x² - 2 = (x + √2)(x - √2) is irreducible over the rational numbers but reducible over the real numbers.

Example: The complex number 4i is a zero of f(x) = x⁴ + 13x² - 48.

Find the remaining zeros of f(x), and write it in its linear factorization.

Since 4i is a zero, then -4i is also a zero so

(x-4i)(x+4i) = x² - (4i)²

= x² - 16i²

= x² + 16

therefore a factor of f(x) is x² + 16

f(x) = x⁴ + 13x² - 48
= (x² + 16)(x² - 3)
= (x + 4i)(x – 4i)(x + √3)(x - √3)

So the zeros are:
x = {-4i, 4i, -√3, √3}

Homework: Pg. 170/ 57, 65, 69
*Pg. 187/ 1 – 5odd, 9 - 19odd, 33, 37, 41, 45, 53, 55, 61, 65

Precalculus 2.4 Complex Numbers

2.4 Complex Numbers

I. A complex number is written in standard form: a + bi
where a ε set of real numbers
and
bi is a pure imaginary number

Therefore a + bi is an imaginary number

√(-1) = i
(√(-1))² = i² = -1
(√(-1))³ = i³ = -i
(√(-1))⁴ = (√(-1))²(√(-1))² = (-1)(-1) = 1
(√(-1))⁰ = 1

Therefore you can see the pattern that:

i⁰ = 1
i ¹= √(-1) = i
i² = -1
i³ = -i
i⁴ = 1
so i⁵ = i
i⁶ = -1
i⁷ = - i
i⁸ = 1

so what would i⁶³ = ?

i⁶⁰i³
i⁶⁰ = 1 and so therefore
i⁶⁰i³ = (1) i³ = - i

B. Additive Identity is zero
what do you add to a number to get that number? of course, zero so

a + bi + ________ = 0

______ = -a - bi

C. Adding and Subtracting complex Numbers:

Example 1: (3 + 4i) + (7 + 2i)
= (3 + 7) + (4i + 2i) by grouping like terms
= 10 = 6i

Example 2: (3 + 4i) - (7 + 2i)
= (3 - 7) + (4i - 2i)
= (-4) + (2i)
= -4 + 2i

D. Multiplying Complex Numbers
Example 1: (3 + 4i)(7 + 2i)
= (3)(7) + (3)(2i) + (4i)(7) + (4i)(2i)
= 21 + 6i + 28i + 8i²
= 21 + 34i + 8(-1)
= 21 + 34i - 8
= 13 + 34i

E. Dividing Complex Numbers - multiply by the complex conjugate

Given "a + bi", the complex conjugate would be "a - bi"

Example 1: complex #1

but recall that "a + bi" has to be in standard form so...

29/53 + 22i/53 would be the answer!

F. Applications:
1. Fractal Geometry

Mandelbrot Set: to draw this, consider: Sequence of numbers

c, c² + c, (c² + c)² + c, [(c² + c)²+ c ]² + c, ...

For some values, it is Bounded, which means that all elements in the sequence are less than some fixed number N. Therefore, complex number "c" is in the Mandelbrot Set.

For other values, it is Unbounded, which means that the elements in the sequence become infinitely large. Therefore, complex number "c" is not in the Mandelbrot Set.

Check out this website:

http://mathworld.wolfram.com/MandelbrotSet.html

Example 1: Let's let c = -1

So the sequence would be:

-1, (-1)² + (-1) = 0, (0)² + (-1) = -1, (-1)² + (-1) = 0

or -1, 0, -1, 0

so this is bounded!

Example 2: Let's let c = - i

-i, (-i)² + (-i) = -1 + -i, (-1 -i)² + (-i) = -3i, (-3i)² + (-i) = -9 - i

or -i, -1 - i, -3i, - 9 - i

so this is not bounded.

2. Impedance - the opposition to current in an electrical circuit.

Equation of 2 pathways:

1/z = 1/z₁ + 1/z₂

where

z₁ is the impedance of pathway 1

z₂ is the impedance of pathway 2

Precalculus 2.3 Real Zeros of Polynomial Functions

Precalculus 2.3a

I. Long Division of Polynomials:

A. Recall: When you did long division

Example 1:

625/5 =

As you can see, 5 goes into 625 perfectly and there is no remainder so this means that 5 is a factor of 625.

Example 2:
625/4 =


As you can see, 4 does not go into 625 perfectly, there is a remainder of 1 so we would write the answer like

156 + 1/4

You write the answer or quotient plus the remainder over the divisor

B. Long Division of Polynomials:

Now try doing the same concept only using Polynomials:

Example 3:


So therefore since (x - 4) is a factor so

x - 4 = 0
x = 4

Therefore (x - 4) is a factor and since there is not a remainder, then (x - h) is a factor and x = h.

Example 4:

Therefore (x + 2) is a factor and since there is not a remainder, then (x - h) is a factor and x = h,
then x = -2

Example 5:




Since when you divide by (x + 1), it has a remainder, it is not a factor and x = -1 is not a zero.

Example 6:





So again, since (x - 2) is not a factor, x = 2 is not a zero.

Precalculus 2.3b Synthetic Division:
To divide ax³ + bx² + cx + d by (x - k), use the following pattern:

Example 1: (5x² - 17x - 12) / (x - 4)

Therefore, (5x² - 17x - 12) /(x - 4) = 5x + 3

since there is not a remainder!

Example 2: (x⁴ + 5x³ + 6x² - x - 2) / (x + 2)

Again, since the remainder is 0,

(x⁴ + 5x³ + 6x² - x - 2) / (x + 2) =

x³ + 3x² - 1

Example 3: (4x³ - 13x + 10) / (x + 1)

Since there is not an x², we put zero in that place:

Since 19 is the remainder:

(4x³ - 13 x + 10) / (x + 1) = 4x² - 4x - 9 + 19/(x + 1)

C. The Remainder Theorem: If a polynomial f (x) is divided by (x - k), the remainder is r = f (x)

Recall from Example 3, we saw the remainder = 19 so...

f (-1) = 4(-1)³ - 13 (-1) + 10= 4 (-1) + 13 + 10= -4 + 13 + 10= 19

this gives the same answer!!

Let's try another:

Example 4: (4x³ - 13x + 10) / (x - 2)

f(2) = 4(2)³ - 13(2) + 10 = 16, again, the same answer!

D. The Factor Theorem:
A polynomial f(x) has a factor (x - k) if and only if f(k) = 0.

In Summary:

The remainder "r", obtained in the synthetic division of f (x) by (x - k) provides the following information:
1. the remainder "r" gives the value of "f" at x = k. That is r = f (k)

2. If r = 0, then (x - k) is a factor of f (x)

3. If r = 0, then (k, 0) is an x-intercept of the graph of f (x).

II. The Rational Zero Test:
relates the possible rational zeros of a polynomial

Rational zero = p/q

where "p" and "q" have no common factors other than 1, "p" is a factor of the constant a₀ and q is a factor of the leading coefficient a_n
polynomial:

f(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₂x² + a₁x + a₀

Possible rational zeros = (factors of constant)/(factors of leading coefficient)

Example 1: f (x) = 4x⁵ - 8x⁴ - 5x³ + 10x² + x - 2

So the possible rational zeros = (1, -1, 2, -2) / (1, -1, 2, -2, 4, -4) so

The possible zeros are : {1, -1, 1/2, -1/2, 1/4, -1/4, 2, -2}

Looking at the graph of the function, we see that the real zeros are:
(-1, 0), (-1/2, 0), (1/2, 0), (1, 0), and (2, 0)

You can also use synthetic Division to eliminate possible zeros:
f (x) = 4x⁵ - 8x⁴ - 5x³ + 10x² + x - 2

try x = -1,

so f(x) = (x + 1)(4x⁴ - 12x³ + 7x² + 3x - 2)


so f (x) = (x - 1)(x + 1)(4x³ - 8x² - 1x + 2)


Therefore f(x) = (x - 1)(x + 1)(x -2)(4x² - 1)

f(x) = (x -1)(x + 1)(x -2)(2x + 1)(2x - 1)

so x = -1, x = 1, x = 2, x = -1/2, x = 1/2

so you can see they are the same answers.

III. Third Test for zeros:

A real Number "b" is an upper bound for the real zeros of "f" if no zeros are greater than "b". Similarily, "b" is a lower bound if no real zeros of "f" are less than "b"

Example 1: f(x) = x⁴ - 4x³ + 15

1. degree is even

2. Leading coefficient is positive

so it rises to the left and rises to the right

3. points (-3, 204), (-2, 63), (-1, 20), (0, 15), (1, 12), (2, -1), (3, -12), (4, 15)

So from the points we chose, the lower bound will be 1 and the upper bound is 4

We can use synthetic division to verify these thoughts:

Let x = -1, you can see that the remainder is 20 and the last row is:

1, -5, 5, -5, 20

Let x = 4, you can see that the remaider is 15 and the last row is:

1, 0, 0, 0, 15

Let x = 1, you can see that the remainder is 12 and the last row is:

1, -3, -3, -3, 12

Therefore we can conclude, given x = c:

1. If c is greater than zero and each number in the last row is either positive or zero, "c" is an upper bound for the real zeros of "f".

2. If c is less than zero and each number in the last row are alternately positive and negative, (zero counts as either positive or negative ), "c" is a lower bound for the real zeros of "f".

Therefore -1 is a lower bound and 4 is an upper bound.