Thursday, 25 May 2017

Wednesday, May 24

In class we worked on converting functions into rational form.

Assignment: 9.8 on p. 555: 1-8, 11-12

Here is an example problem, similar to the one worked in class.

Let's analyze the function given by the equation below. We will try to work out its various characteristics and sketch a graph before resorting to computer graphing.

y = dfrac{x-1}{(x+1)(x - 2)} + dfrac{2x - 1}{x^2 - 4}

To obtain a common denominator we multiply the first term by (x+2)/(x+2) and the second term by (x + 1)/(x+1)

y = dfrac{(x - 1)(x + 2)}{(x + 1)(x - 2)(x + 2)} + dfrac{(2x - 1)(x + 1)}{(x + 1)(x - 2)(x + 2)}

Now we can simplify the expressions in the numerators and re-write as a single fraction.

y = dfrac{x^2 + x - 2 + 2x^2 + x - 1}{(x +1)(x - 2)(x + 2)}

Which simplifies to

y = dfrac{3x^2 + 2x - 3}{(x + 1)(x - 2)(x + 2)} or y = dfrac{3x^2 + 2x - 3}{x^3 + x^2 - 4x - 4}

We note that 3x^2 + 2x - 3, the numerator, cannot be factored. So, later on as we analyze the properties of this function we will need to solve 3x^2 + 2x - 3 = 0 to determine the value of any x-intercepts.

Now that the function is in rational form, we can analyze its properties.

We note that there are no holes in the function. Because there are no factors common to both the numerator and denominator. i.e. (x + 2)/(x + 2). These common factors would work out to essentially multiplying the function by 1 for all values of $x$ except for when $x + 2 = 0$ at $x = -2$. For that single value $(x + 2)/(x + 2)$ becomes undefined and we get a hole in the graph. Anyway, since this is not the case, our function has no holes.

The function has vertical asymptotes at

x = -1, x = 2 and x = -2

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We can tell from the numerator that the function will have x-intercepts where the numerator evaluates to zero, so we solve 3x^2 + 2x - 3 = 0 to get

x = dfrac{-2 pm sqrt{2^2 - 4(3)(-3)}}{2(3)}

x = dfrac{-2 pm sqrt{40}}{6} = dfrac{-2 pm 2sqrt{10}}{6} = dfrac{-1 pm sqrt{10}}{3}  approx -1.387,0.72

So, we note the two x-intercepts.

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In addition, we note that this function has denominator of greater degree than the numerator, so as x approaches pm infty the denominator will overpower the numerator and the value of the function will approach 0, creating horizontal asymptotes at y = 0

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Through point testing and some more intuition we can identify the location of the 4 sections of this function and sketch a graph

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Thursday, 18 May 2017

Thursday, May 18

In class: we studied the definition of /rational functions/ and looked at the graph of f(x) = 1/x and transformations of this parent function.

For example the function f(x) = 1/x

When transformed to g(x) = 3/(x-4) + 1 incurs a vertical stretch by factor 3 and horizontal shift of +4 to the right and vertical shift + 1. The asymptotes of the function are transformed with the function, so the asymptotes y = 0 and x = 0 of the parent function are vertically stretched (no effect) then vertically shifted (y -> 3fy + 1) and horizontally translated (x -> x + 4), so the x = 0 vertical asymptote moves right to x = 4.

Also, our nice point (1,1) moves 4 to the right and takes a vertical stretch by factor 3 followed by a vertical shift + 1, so (1 + 4, 1*3+1) = (5, 4)

Assignment: 9.6 on p. 536: 1-8, 10-12

Reminder: SBAC testing in Ms. Mac’s room tomorrow during 6th period.

Tuesday, 16 May 2017

Tuesday, May 16

In class we reviewed various forms of conics (ellipses, hyperbola, parabola) and learned about the general form of a quadratic

Ax^2 + bxy + Cy^2 + Dx + Ey + F = 0

Today's problems ask students to convert between this general form and the standard form of various other conic sections.

Assignment: 9.5 on p. 531: 1-8, 10, 13-16

Monday, 8 May 2017

Monday, May 8

In class: notes on the locus/distance definition of a parabola and how to find the equation of a parabola given the equation and position of the directrix and focus.

Assignment: Read 9.3 and complete 1-9 on p. 511.

Tuesday, 2 May 2017

Thursday, 27 April 2017

Thursday, April 27

In class we solved a "smart pilot" problem in which we determine the proper bearing for a plane to take given the plane's airspeed, a wind vector and the distance and bearing between the origin and destination.


1) Chapter 8 Review on p. 483: complete 1-11

2) Graded Problem write-up

Smart Pilot, MYP Criteria D Problem Write Up               

  • Question: How would a pilot plot a rudimentary course between two points using trigonometry to determine the correct bearing and flight duration?
  • Find two cities and determine the distance and bearing from one to the other (as the crow flies)
  • Find the airspeed of a plane
  • Find a wind speed and bearing that you think would be reasonable (may not be in a nice line with your plane)
  • Determine the bearing which the pilot should follow, the duration of the flight
  • Present your work and explanation [Graded on Criteria D] Due on Tuesday

Tuesday, 25 April 2017

Saturday, 22 April 2017

Friday, April 21, Law of Sines

In class: notes on Law of Sines — using it to solve for missing sides and angles in a general triangle, the ambiguous case SSA. Ambiguous only when the angle is acute, and the non-adjacent side is shorter than the adjacent side, and greater than the length given by assuming that the triangle is right and solving for the side length using sine/cosine/tangent ratios.

1) 8.6 on p. 472: 1-9, 15
2) Read 8.7 on Law of Cosines
3) Complete 8.5/8.6 practice problems from packet

Tuesday, 18 April 2017

Tuesday, April 18

In class gravity and projectile motion discussed. Students should be able to model the path of an object subject to the force of gravity and initial velocity and angle (ignoring friction/air resistance) and do so using both parametric and non-parametric equations.


1) 8.3/8.4 worksheet

2) 8.5 on p. 464: 2-12 even, 13-16. Also, note that you will need the gravity table on p. 460

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3) Read the section before/during or after you complete your assignment

Friday, 14 April 2017