Category Archives: MAC1105

1.3 Complex Numbers, Complex Numbers

Powers of i

June 5, 2017 admin

Example: Simplify the power of i.

i^-59

Solution:

i^-59

Rewrite the exponential expression with negative exponents by using the reciprocal property.

= $1/{i^59}$

Simplify the i^59 by dividing 59 by 4. The remainder will tell you which position in the pattern you are in. 59 divided by 4 is 14 with a reminder of 3.

= $1/{i^3}$

= 1/{-i}

= 1/{-i} {i /i}

= $i/{-i^2}$

= i/{-(-1)}

= i/{1}

1.2 Applications of Linear Equations, Applications of Linear Equations

Consecutive Numbers

May 23, 2017 admin

Here is a handout about consecutive numbers.

MAT0024_2.5CW Consecutive

1.2 Applications of Linear Equations, Translating Phrases

Translating Phrases

May 23, 2017 admin

Here is a handout on translating phrases from words into mathematical expressions and equations. Solutions are provided on last page.

MAT0024_1.1CW Translating Phrases

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

Graphing an Exponential Equation by Transformations

April 29, 2017 admin

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.

x	y
-2	2^(-2)=1/4
-1	2^(-1)=1/2
0	2^0=1
1	2^1=2
2	2^2=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.

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

Quadratic Function: General Form

March 30, 2017 admin

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.

2.1 Graphing, Distance

Using the Distance Formula

March 25, 2017 admin

The Distance Formula

Suppose A is ( x_1, y_1) and B is ( x_2, y_2)

The distance between points A and B is given by the following formula.

$d=sqrt {(x_2-x_1)^2+ (y_2-y_1)^2}$

Example: Find the y-coordinate of the points that are 5 units away from the point (3,3) that have an x-coordinate of -1.

$5=sqrt {(-1-3)^2+ (y-3)^2}$

$5=sqrt {(-4)^2+ (y-3)^2}$

$5=sqrt {16+ (y-3)^2}$

$5^2=(sqrt {16+ (y-3)^2})^2$

25=16+ (y-3)^2

25-16=16-16+ (y-3)^2

9= (y-3)^2

$sqrt{9}= sqrt{(y-3)^2}$

pm 3= y-3

3 pm 3= y

3 + 3= y or 3 - 3= y

6= y or 0= y

2.1 Graphing, 3.6 Mathematical Models: Building Functions, Distance

Finding the Distance

March 25, 2017 admin

The Distance Formula

Suppose A is ( x_1, y_1) and B is ( x_2, y_2)

The distance between points A and B is given by the following formula.

$d=sqrt {(x_2-x_1)^2+ (y_2-y_1)^2}$

Example: Find the distance between points A and B.

Point A is ( 4, -2) and point B is ( -8, 3)

$d=sqrt {(-8-4)^2+ (3-(-2))^2}$

$d=sqrt {(-12)^2+ (3+2)^2}$

$d=sqrt {144+ (5)^2}$

d=sqrt {144+ 25}

d=sqrt {169}

d=13

2.1 Graphing, Midpoint

Finding the Midpoint

March 25, 2017 admin

The Midpoint Formula

Suppose A is ( x_1, y_1) and B is ( x_2, y_2)

The midpoint of the line segment joining points A and B is given by the following formula.

$( {x_1+x_2}/2, {y_1+y_2}/2)$

Example: Find the midpoint of a line segment joining points A and B.

Point A is ( 4, -2) and point B is ( -8, 3)

The x coordinate of the midpoint can be found by averaging the x-values of point A and B. The y-coordinate of the midpoint can be found by averaging the y-values of point A and B.

( {4+(-8)}/2, {-2+3}/2)