Mathematics for Technology I (Math 1131) Durham College, Mathematics
Free • 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 x-y 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.

• Geometry

This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.

Mathematics for Technology I (Math 1131)

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 x-intercepts), vertex point, and y-intercept. 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 x-coordinate 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:

1.    x1 = -0.4188
2.    x2 = -3.5811

To take the average of any two numbers, you sum them and divide by 2:

$\frac{{x}_{1}+{x}_{2}}{2}=\frac{–0.4188+\left(–3.5811\right)}{2}=–2\phantom{\rule{0ex}{0ex}}$

–2 is the x-coordinate of the vertex and the axis of symmetry. The y-coordinate is found by substituting this value back into the original equation:

∴ the vertex is at (–2, 5), and the x-intercepts 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 y-intercept; this is done by setting x = 0:

∴ the y-intercept is (0, –3). Here’s what a sketch would look like (below). One thing to be mindful of is the a-coefficient 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 x-axis (therefore, no solutions), you’ll need to use a special formula to find the vertex.

$h=–\frac{b}{2a}\phantom{\rule{0ex}{0ex}}$
• Where h represents the x-coordinate 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 y-coordinate of the vertex.

Question:   Sketch the following function:

$f\left(x\right)=2.74{x}^{2}–3.12x+5.38\phantom{\rule{0ex}{0ex}}$

Solution:   Here’s what we know about the function:

1. a = 2.74 ; b = –3.12 ; c = 5.38
2. Parabola opens up because 2.74 is positive.
3. There are no x-intercepts because the discriminant is less than 1.
• $\sqrt{{\mathbit{b}}^{\mathbf{2}}\mathbf{–}\mathbf{4}\mathbit{a}\mathbit{c}}={\left(–3.12\right)}^{2}–4\left(2.74\right)\left(5.38\right)<0\phantom{\rule{0ex}{0ex}}$
4. The x-coordinate of the vertex is:
• $h=–\frac{b}{2a}=\frac{–\left(–3.12\right)}{2\left(2.74\right)}=\mathbf{0}\mathbf{.}\mathbf{569}\phantom{\rule{0ex}{0ex}}$
• ∴ the y-coordinate is: f( 0.569 ) = 2.74 ( 0.569 )² – 3.12 ( 0.569 ) + 5.38 ⇒ 4.49
5. The y-intercept occurs when x = 0, f( 0 ) = 2.74 ( 0 )² – 3.12 ( 0 ) + 5.38 ⇒ (0, 5.38)
6. The graph looks like: 