**Assignment:**12.2 on p. 628: 1-16, 21

# Month: May 2017

#### Friday, 26 May 2017

#### 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.

To obtain a common denominator we multiply the first term by and the second term by

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

Which simplifies to

or

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

**in the function. Because there are no factors common to both the numerator and denominator. i.e. . 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**

*there are no holes**no holes*.

The function has vertical asymptotes at

and

We can tell from the numerator that the function will have x-intercepts where the numerator evaluates to zero, so we solve to get

=

So, we note the two x-intercepts.

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

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

# Wednesday, May 24

Students took the Similarity test (chapter 11) on Friday. This week we have learned the right-triangle definition of the trigonometric functions sine, cosine and tangent. As of now these functions take as input one of the *non-right* angles in a right triangle and their output is a specific ratio of sides in *all such similar right triangles*.

So, based on this definition, if that means that in any right triangle with a angle, the ratio of * *is approximately .57*.* In the figure below this means that *.*

**Assignments**

Monday: No HW

Tuesday: 12.1 on p. 620-624: 4-9, 14-22, 27

Wednesday

1. "Understanding sine/cosine/tangent and their inverses" w/s

2. (3rd Period) 12.1 worksheet

2. (5th Period) 12.2 worksheet

# Wednesday, May 24

* The distance formula (section 8.1)

* Using coordinates to prove that segments are congruent, parallel or perpendicular (8.2)

* Using congruence, perpendicularity and parallel properties to justify conjectures about shapes. (8.3)

* Writing and analyzing vertical shifts in linear or exponential functions (these have the form f(x) = g(x) + k).

Past assignments

Friday — 8.4 Task + Go (in class) completed on whiteboards, not necessarily in notebook.

Monday — 8.5 Task + RSG

Tuesday — 8.6 Task (Part 1/2/3 all) + 8.6 Set

**Today’s assignment: Algebra & Geometry Review worksheet**

#### Thursday, 18 May 2017

# Thursday, May 18

In class we spent time reviewing for tomorrow’s Similarity (Ch. 11) Test.

Assignment: Chapter 11 Review on p. 624: 1-21 (pick 12 problems) & be ready for your test tomorrow.

# 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.

# Thursday, May 18

**Today's class.**Review of 8.3 distance problems, graphing video activity, no hw.

#### Wednesday, 17 May 2017

# Wednesday, May 17

In class problem solving to prepare for Friday’s test on Similarity (Ch. 11).

Assignment: Challenge Problems + 11.5/11.6 PYS worksheet.

# Wednesday, May 17

These problems deal with proving congruence, perpendicularity, and parallelism using coordinate relationships. So for example, showing that the shape below is a rectangle by demonstrating that the slope of each pair of adjacent sides is perpendicular. We demonstrate perpendicularity by computing the slopes of the lines say slope(BC) = -(2/6) = -1/3 and the slope(AB) = 3/1 = 3. Since -1/3 and 3 are negative reciprocals we can see that sides AB and BC are perpendicular, and thus the angle at B is a right angle.

#### Tuesday, 16 May 2017

# Tuesday, May 16

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