# Thursday, May 25

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

# 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.

$\dpi{300}\inline y = \dfrac{x-1}{(x+1)(x - 2)} + \dfrac{2x - 1}{x^2 - 4}$

To obtain a common denominator we multiply the first term by $\dpi{300}\inline (x+2)/(x+2)$ and the second term by $\dpi{300}\inline (x + 1)/(x+1)$

$\dpi{300}\inline y = \dfrac{(x - 1)(x + 2)}{(x + 1)(x - 2)(x + 2)} + \dfrac{(2x - 1)(x + 1)}{(x + 1)(x - 2)(x + 2)}$

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

$\dpi{300}\inline y = \dfrac{x^2 + x - 2 + 2x^2 + x - 1}{(x +1)(x - 2)(x + 2)}$

Which simplifies to

$\dpi{300}\inline y = \dfrac{3x^2 + 2x - 3}{(x + 1)(x - 2)(x + 2)}$ or $\dpi{300}\inline y = \dfrac{3x^2 + 2x - 3}{x^3 + x^2 - 4x - 4}$

We note that $\dpi{300}\inline 3x^2 + 2x - 3$, the numerator, cannot be factored. So, later on as we analyze the properties of this function we will need to solve $\dpi{300}\inline 3x^2 + 2x - 3 = 0$ 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. $\dpi{300}\inline (x + 2)/(x + 2)$. 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

$\dpi{300}\inline x = -1, x = 2$ and $\dpi{300}\inline x = -2$

We can tell from the numerator that the function will have x-intercepts where the numerator evaluates to zero, so we solve $\dpi{300}\inline 3x^2 + 2x - 3 = 0$ to get

$\dpi{300}\inline x = \dfrac{-2 \pm \sqrt{2^2 - 4(3)(-3)}}{2(3)}$

$\dpi{300}\inline x = \dfrac{-2 \pm \sqrt{40}}{6}$ = $\dpi{300}\inline \dfrac{-2 \pm 2\sqrt{10}}{6} = \dfrac{-1 \pm \sqrt{10}}{3}$  $\dpi{300}\inline \approx -1.387,0.72$

So, we note the two x-intercepts.

In addition, we note that this function has denominator of greater degree than the numerator, so as $\dpi{300}\inline x$ approaches $\dpi{300}\inline \pm \infty$ the denominator will overpower the numerator and the value of the function will approach $\dpi{300}\inline 0$, creating horizontal asymptotes at $\dpi{300}\inline y = 0$

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 $\dpi{300}\inline \sin(35^\circ) \approx .57$ that means that in any right triangle with a $\dpi{300}\inline 35^\circ$ angle, the ratio of $\dpi{300}\inline \dfrac{\text{opposite side}}{\text{adjacent side}}$ is approximately .57. In the figure below this means that $\dpi{300}\inline \dfrac{AC}{BC} \approx .57$.

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

We are finished with Module 8. Today we spent class time reviewing Module 8 skills. There is a test covering

* 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, 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, 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, May 16

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

$\dpi{300}\inline Ax^2 + bxy + Cy^2 + Dx + Ey + F = 0$

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