 55 lessons
 1 quizzes
 10 week duration

Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.

Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an xy plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.

Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.

Vector Analysis

Introduction to Algebra

Factoring

Solving Equations

Functions and Graphs

Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
 Identify, measure, and calculate different types of straight lines and angles
 Calculate the interior angles of polygons
 Solve problems involving a variety of different types of triangles
 Calculate the area of a variety of different types of quadrilaterals
 Solve problems involving circles
 Calculate the areas and volumes of different solids

Introduction to Statistics
Graphing Quadratic Functions
Graphing a quadratic function without a table of values isn’t as simple as linear functions. This process involves knowing the parabola’s direction of opening, and finding its solutions (also called roots or xintercepts), vertex point, and yintercept. These features are shown below:
Sketching Quadratics with Roots
Recall that to graph a quadratic function, we could use a table of values with a specified domain (3 to 3, for example). However, this isn’t practical all the time. You learned previously that you can find the roots of a quadratic function using the quadratic formula, and that it can have up to 2 roots. If two roots are present, the xcoordinate of the vertex can be found by taking the average of the two roots because all parabolas are symmetrical and its axis of symmetry lies on the vertex point. For example, if the quadratic function is f(x) = −2x² − 8x − 3, the roots will be:
 x_{1} = 0.4188
 x_{2} = 3.5811
To take the average of any two numbers, you sum them and divide by 2:
$\frac{{x}_{1}+{x}_{2}}{2}=\frac{\u20130.4188+\left(\u20133.5811\right)}{2}=\u20132\phantom{\rule{0ex}{0ex}}$–2 is the xcoordinate of the vertex and the axis of symmetry. The ycoordinate is found by substituting this value back into the original equation:
$f\left(\u20132\right)=\; 2{\left(\u20132\right)}^{2}\u20138\left(\u20132\right)\u20133\phantom{\rule{0ex}{0ex}}f(\u20132)=5\phantom{\rule{0ex}{0ex}}$∴ the vertex is at (–2, 5), and the xintercepts are at (0.4188, 0) and (3.5811, 0). You can use these three points to create a sketch of the parabola. In fact, you can take this a step further by easily finding the yintercept; this is done by setting x = 0:
$f\left(0\right)=\; 2{\left(0\right)}^{2}\u20138\left(0\right)\u20133\phantom{\rule{0ex}{0ex}}f\left(0\right)=\overline{)\u20132{\left(0\right)}^{2}}\overline{)\u20138\left(0\right)}\u20133\phantom{\rule{0ex}{0ex}}f\left(0\right)=\u20133\phantom{\rule{0ex}{0ex}}$∴ the yintercept is (0, –3). Here’s what a sketch would look like (below). One thing to be mindful of is the acoefficient determines if a parabola opens up or down. Give that it’s negative in this function, we see a parabola opening down, otherwise it would be up like a smile.
Sketching Quadratics without Roots
In case you’re provided a quadratic that doesn’t cross the xaxis (therefore, no solutions), you’ll need to use a special formula to find the vertex.
$h=\u2013\frac{b}{2a}\phantom{\rule{0ex}{0ex}}$ Where h represents the xcoordinate of the vertex. Note that this formula can also be used if roots exist.
Once you have found h, you substitute this value into the equation to find the corresponding ycoordinate of the vertex.
Question: Sketch the following function:
$f\left(x\right)=2.74{x}^{2}\u20133.12x+5.38\phantom{\rule{0ex}{0ex}}$Solution: Here’s what we know about the function:
 a = 2.74 ; b = –3.12 ; c = 5.38
 Parabola opens up because 2.74 is positive.
 There are no xintercepts because the discriminant is less than 1.
 $\sqrt{{\mathit{b}}^{\mathbf{2}}\mathbf{\u2013}\mathbf{4}\mathit{a}\mathit{c}}={\left(\u20133.12\right)}^{2}\u20134\left(2.74\right)\left(5.38\right)<0\phantom{\rule{0ex}{0ex}}$
 The xcoordinate of the vertex is:
 $h=\u2013\frac{b}{2a}=\frac{\u2013\left(\u20133.12\right)}{2\left(2.74\right)}=\mathbf{0}\mathbf{.}\mathbf{569}\phantom{\rule{0ex}{0ex}}$
 ∴ the ycoordinate is: f( 0.569 ) = 2.74 ( 0.569 )² – 3.12 ( 0.569 ) + 5.38 ⇒ 4.49
 The yintercept occurs when x = 0, f( 0 ) =
2.74 ( 0 )²– 3.12 ( 0 )+ 5.38 ⇒ (0, 5.38) The graph looks like: