Category Archives: MAC1140

3.1 Functions, 3.1 Relations and Functions, 4.6 Rational Functions and Their Graphs, 5.4 Properties of Rational Functions

Finding Domain: Rational Function

Example:  Classify the function as a polynomial function, rational function, or root function, and then find the domain.  Write the domain interval notation and set builder notation.

h(x) = {x^2+8}/{x^2+x-20}

Solution:

Classify the Function

Formal Definition
Practical Way of Identifying
Polynomial Function

A polynomial function is a function of the form

f(x)=a_n x^n+a_{n-1} x^{n-1}+...+a_1 x+a_0

where n is a non-negative integer {0, 1, 2, 3, 4, …} and the coefficients a_n , a_{n-1,...,a_1,a_0 are from the real numbers.
Look for the variables to be in the numerator.  (If there is no fraction at all, they are in the numerator.)  The variable should not be inside a radical or absolute value.  The powers or exponents on the variables should be whole numbers.  Whole numbers come from this list {0, 1, 2, 3, 4, …}
Rational Function

A rational function is a function of the form

f(x)={p(x)}/{q(x)}

where p(x) and q(x) are polynomial functions and q(x) is not equal to zero.
The numerator and denominator are polynomials.  Most functions with variables in the denominator are considered Rational Functions but there are exceptions.
 Root Function (even index)

A root function is a function of the form

where n is an even positive integer greater than or equal to 2.

  The variable is inside or underneath a radical.  The index of the radical is an even number.  {2, 4, 6, 8, …}  The square root is an even index although the index is not written.
 Root Function (odd index)

A root function is a function of the form

where n is an odd positive integer greater than or equal to 2.

 The variable is inside or underneath a radical.  The index of the radical is an odd number.  {3, 5, 7, 9, …}  The cube root is an odd index.

Since the function h(x) = {x^2+8}/{x^2+x-20} has a variable in the denominator and the numerator and denominator are polynomial functions this function is a rational function.

Find the Domain of a Rational Function

Division by zero is undefined. Having a zero as the denominator is equivalent to division by zero thus is also undefined. The rational function is undefined for any value of the variable that gives a zero denominator. Find these values by creating an equation to solve. The equation is the expression in the denominator equal to zero.

Set the denominator equal to zero.
x^2+x-20=0
Solve the equation. This equation is a quadratic equation and can be solved by factoring, completing the square or the quadratic formula. x^2+x-20=0
Solve by factoring.  Factor the expression on one side. x^2+x-20=0
(x+5)(x-4)=0
Use the zero product property and set each factor equal to zero. (x+5)(x-4)=0
x+5=0 or x-4=0
Solve each equation. x+5=0 or x-4=0
x+5-5=0-5 or x-4+4=0+4
x=-5 or x=4

The values -5 and 4 give a zero value in the denominator, make the function undefined and must be excluded from the domain.

In set builder notation, the domain is delim{lbrace}{ x delim{ |} x!=-5 and x!=4}{rbrace}

In interval notation, the domain is (- infty,-5) union (-5,4) union (4,infty)

 

7.3 The Ellipse

Analyze the Equation of an Ellipse

Example: Find the center, foci, and vertices of the ellipse.  Graph the equation.

(x-1)^2/25+(y+1)^2/64=1

Solution:

Finding the center:

There are two standard orientations of ellipse’s, both of them have the center at (h,k).

Horizontal:  (x-h)^2/a^2+(y-k)^2/b^2=1

Vertical: (x-h)^2/b^2+(y-k)^2/a^2=1

For this example the center of the ellipse is (1,-1)

Finding the orientation:

The larger denominator will indicate the orientation.  Since our example has a larger denominator under the y variable the orientation is vertical.  That means the ellipse is longer vertically.

Finding the vertices: (Major Axis)

a is the distance between the center and the vertex along the major or longer axis.  Since our larger denominator is 64 a is 8.  I know this because the in the formula the larger denominator is a^2. If a^2=64 then a=8 .

Since the ellipse is vertically oriented the vertices will be 8 units above and below the center.

This puts the vertices at (1,7) and (1,-9).

Finding the vertices: (Minor Axis)

b is the distance between the center and the vertex along the minor or shorter axis.  Since our smaller denominator is 25 b is 5.  I know this because the in the formula the smaller denominator is b^2. If b^2=25 then b=5 .

Since the ellipse is vertically oriented the vertices along the minor axis or shorter axis will be 5 units left and right of the center.

This puts the vertices at (-4,-1) and (6,-1).

Finding the foci: 

c is the distance between the center and the focus.  You can not find c from the equation.  You must know that a, b, and c are related in a similar formula to Pythagorean theorem.

b^2= a^2-c^2

Fill in a and b, then you will be able to solve for c.

b^2= a^2-c^2

5^2= 8^2-c^2

25= 64-c^2

25-64= 64-64-c^2

-39=-c^2

39=c^2

sqrt{39}=sqrt{c^2}

pm sqrt{39}=c

Since c is a distance we will use the positive solution.  sqrt{39}=c.

Since the ellipse is vertically oriented the foci will be sqrt{39} above and below the center.

This puts the foci at (1,-1+sqrt{39}) and (1,-1-sqrt{39})