Category Archives: MAC1140

3.1 Functions, 3.1 Relations and Functions, Difference Quotient

Difference Quotient: Rational Function

Example:  Find the difference quotient for f(x)={4x}/{x+5}

The Difference Quotient:{f(x+h)-f(x)}/h

Solution:

The Difference Quotient Formula
{f(x+h)-f(x)}/h
Write the difference quotient for the given function
={{4(x+h)}/{(x+h)+5}-{4x}/{x+5}}/h
Use the distributive property
={{4x+4h}/{x+h+5}-{4x}/{x+5}}/h
Simplify the complex fraction by multiplying the numerator and denominator by the common denominator
={{4x+4h}/{x+h+5}-{4x}/{x+5}}/h {{(x+5)(x+h+5)}/1}/ {{(x+5)(x+h+5)}/1}
Distribute the common denominator to each fraction in the numerator.
={{(4x+4h)(x+5)(x+h+5)}/{x+h+5}-{4x(x+5)(x+h+5)}/{x+5}} /{h(x+5)(x+h+5)}
Cancel the common factor
={{(4x+4h)(x+5)}-{4x(x+h+5)}} /{h(x+5)(x+h+5)}
 Multiply the expression in the numerator
={4x^2+20x+4xh+20h-4x^2-4xh-20x} /{h(x+5)(x+h+5)}
 Combine like terms
={20h} /{h(x+5)(x+h+5)}
 Cancel a common h from the numerator and denominator
={20} /{(x+5)(x+h+5)}

 

3.1 Functions, 3.1 Relations and Functions, Difference Quotient

Difference Quotient: Rational Function

Example:  Find the difference quotient for f(x)=1/x^2

The Difference Quotient:{f(x+h)-f(x)}/h

Solution:

The Difference Quotient Formula
{f(x+h)-f(x)}/h
Write the difference quotient for the given function
={1/(x+h)^2-1/x^2}/h
Simplify the complex fraction by multiplying the numerator and denominator by the common denominator
={1/(x+h)^2-1/x^2}/h {{x^2(x+h)^2}/1}/{{x^2(x+h)^2}/1}
Distribute the common denominator to each fraction in the numerator.
={1/(x+h)^2-1/x^2}/h {{x^2(x+h)^2}/1}/{{x^2(x+h)^2}/1}
={{x^2(x+h)^2}/(x+h)^2-{x^2(x+h)^2}/x^2}/{h{x^2(x+h)^2}}
Cancel the common factor
={{x^2(x+h)^2}/(x+h)^2-{x^2(x+h)^2}/x^2}/{h{x^2(x+h)^2}}
={x^2-(x+h)^2}/{h{x^2(x+h)^2}}
Simplify by squaring the binomial
={x^2-(x+h)^2}/{h{x^2(x+h)^2}}
={x^2-(x+h)(x+h)}/{h{x^2(x+h)^2}}
={x^2-(x^2+xh+xh+h^2)}/{h{x^2(x+h)^2}}
Simplify by combining like terms and distributing the negative
={x^2-(x^2+2xh+h^2)}/{hx^2(x+h)^2}
={x^2-x^2-2xh-h^2}/{hx^2(x+h)^2}
 Combine like terms
={-2xh-h^2}/{hx^2(x+h)^2}
 Cancel a common h from the numerator and denominator
={-2x-h}/{x^2(x+h)^2}

The difference quotient for f(x)=1/x^2 is {-2x-h}/{x^2(x+h)^2}

3.5 Graphing Techiques: Transformations, 5.1 Exponential Functions, 6.3 Exponential Functions, Graphing Exponential Functions

Graphing an Exponential Equation by Transformations

Example: For the function below.  Graph using transformations.  Find the y-intercept.  State the horizontal asymptote and the domain and range.

f(x)=-2^{x+2}+2

First we must examine the base function y=2^x

Graph using plotting points.  We can use the standard set of x-values to find ordered pairs.

xy
-22^(-2)=1/4
-12^(-1)=1/2
02^0=1
12^1=2
22^2=4

The graph below shows the points plotted and the line that connects them.  This graph has a horizontal asymptote at y=0.  The domain is (- infty, infty)  and the range is (0, infty)

 

Analyze the transformations.

f(x)=-2^{x+2}+2

The +2 in the exponent shifts the graph left 2 units.

The – in the front of the base reflects the graph over the x-axis.

The +2 next to the base shifts the graph and the horizontal asymptote up two units.

You can see the graph after the transformations.

The horizontal asymptote is y=2.  The domain is (- infty, infty)  and the range is (- infty, 2)

To find the y-intercept we let x=0.

f(0)=-2^{0+2}+2

f(0)=-(2^{2})+2

f(0)=-(4)+2

f(0)=-4+2

f(0)=-2

Thus the y-intercept is (0,-2)

Here is a youtube video with examples.

 

 

9.2 Arithmetic Sequences

Sum of Arithmetic Sequence Application

A quilt is designed in the shape of an equilateral triangle,  5 inches on each side.  Each section of the quilt is in the shape of an equilateral triangle, 1 inch on a side.  The sections are to alternate in color as show in the picture.  How many dark green sections will be required?  How many light green sections will be required?

 

Dark Green Sections

You can use an arithmetic sequence to represent the number of triangles in each row since the number of triangles increases by 1 in the next consecutive row.

The formula for the nth term of an arithmetic sequence is below.

  a_n = a _1+ (n-1)d

d is the common difference and   a_1 is the first term in the sequence.

For the dark green sections, there is one dark green triangle in the first row and the number of dark green triangles increases by 1 in the next row.  This means   a_1=1 and   d=1

Using the formula the nth term of the arithmetic sequence is the following.

  a_n = a_1 +(n-1)d

  a_n = 1 + (n-1)(1)

  a_n = 1 + n-1

  a_n = n

 

Now that the nth term of the sequence has been established we will find the sum of the rows.  Each term of the sequence represents the number of dark green triangles are in that row.  Summing the terms of the sequence is finding the total number of dark green triangles.

The formula for the sum of the first n terms of an arithmetic sequence is below.

  S_n=sum{i=1}{n}{{[}{{a_1+(i-1)d}}}{]}= n/2(a_1+a_n)

There are five rows of dark green triangles.  The sum the first five rows of dark green triangles is represented by the symbols below.

  sum{i=1}{5}{i}

To use the formula we need the first term, the last term and n.  The first term in the sequence is 1 (1 dark green triangle in the first row.  The last term in the sequence is 5 (5 dark green triangles in the last row.  Since there are a total of 5 rows, n is 5.

  sum{i=1}{5}{i}=n/2(a_1+a_n)

  sum{i=1}{5}{i}= 5/2 (1+5)

  sum{i=1}{5}{i}= 5/2 (6)

  sum{i=1}{5}{i}= 30/2

  sum{i=1}{5}{i}= 15

The number of dark green sections required is 15.

Light Green Sections

You can use an arithmetic sequence to represent the number of triangles in each row since the number of triangles increases by 1 in the next consecutive row.

The formula for the nth term of an arithmetic sequence is below.

  a_n = a _1+ (n-1)d

d is the common difference and   a_1 is the first term in the sequence.

For the light green sections, there are no light green triangles in the first row and the number of light green triangles increases by 1 in the next row.  This means   a_1=0 and   d=1

Using the formula the nth term of the arithmetic sequence is the following.

  a_n = a_1 +(n-1)d

  a_n = 0 + (n-1)(1)

  a_n = 0 + n-1

  a_n = n-1

Now that the nth term of the sequence has been established we will find the sum of the rows.  Each term in the sequence represents the number of light green triangles in that row.  The sum of the sequence will give us the total number of light green triangles.

The formula for the sum of the first n terms of an arithmetic sequence is below.

  S_n=sum{i=1}{n}{{[}{{a_1+(i-1)d}}}{]}= n/2(a_1+a_n)

There are five rows of dark green triangles.  The sum the first five rows of light green triangles is represented by the symbols below.

  sum{i=1}{5}{i-1}

To use the formula we need the first term, the last term and n.  The first term in the sequence is 0 (0 light green triangles in the first row.  The last term in the sequence is 4 (4 light green triangles in the last row.  Since there are a total of 5 rows, n is 5.

  sum{i=1}{5}{(i-1)}=n/2(a_1+a_n)

  sum{i=1}{5}{(i-1)}= 5/2 (0+4)

  sum{i=1}{5}{(i-1)}= 5/2 (4)

  sum{i=1}{5}{(i-1)}= 20/2

  sum{i=1}{5}{(i-1)}= 10

The number of light green sections required is 10.

 

4.1 Quadratic Functions, 4.3 Quadratic Functions and Their Properties, Graphing Quadtratic Functions

Quadratic Function: General Form

Example: Rewrite the given quadratic function in standard form by completing the square.  Then state the domain, range, vertex, x-intercepts, y-intercept, the orientation (opens up or opens down), and the axis of symmetry.  Finally graph the function.

f(x)=-x^2-2x+8

Solution:

Rewrite function in standard form.

f(x)=-x^2-2x+8

f(x)=-(x^2+2x)+8

f(x)=-(x^2+2x+1)+8+1

f(x)=-(x+1)^2+9

Here is a youTube video that demonstrations the process.

Find the vertex.

When the quadratic function is written in standard form you can identify the vertex as (h,k).

f(x)=a(x-h)^2+k

f(x)=-(x+1)^2+9

The vertex of the quadratic function is (-1,9).

Find the orientation.

The leading coefficient of the quadratic function is negative so the parabola opens down.

Find the y-intercept.

To find a y-intercept let x=0.

f(0)=-(0+1)^2+9

f(0)=-(1)^2+9

f(0)=-1+9

f(0)=8

The y-intercept is (0,8).

Find the x-intercept.

To find an x-intercept let y=0 or f(x)=0.

0=-(x+1)^2+9

0-9=-(x+1)^2+9-9

-9=-(x+1)^2

{-9}/{-1}={-(x+1)^2}/{-1}

9=(x+1)^2

sqrt{9}=sqrt{(x+1)^2}

pm 3=x+1

-1 pm 3=x+1-1

-1 pm 3=x

-1 + 3=x or -1 - 3=x

2=x or -4=x

Find the axis of symmetry.

The axis of symmetry is a vertical line that passes through the vertex.  Since the vertex is (-1,9) the axis of symmetry is x=-1.

Graph the function.

Plot the intercepts and the vertex.

Since the vertex is the highest point we can draw the parabola using the peak at the vertex.

Find the domain and range.

The domain is (- infty, infty).

The range is (  - infty, 9 {}{]}.

Here a few video examples of analyzing the graph in general form.

2.2 Lines, 2.3 Lines, 4.1 Linear Functions and Their Properties, Graphing with Slope Intercept Form

Graphing Linear Equations with Slope Intercept Form

Example: Graph the linear equation using the slope and the y-intercept. (the slope is positive and fractional)

Example: Graph the linear equation using the slope and the y-intercept. (positive and negative fractional slopes)

Example: Graph the linear equation using the slope and the y-intercept. (equation written in standard form)