# Blackout Poem – “Death of a Salesman”

Blackout Poem

Eerily quiet.

Intuited what was to come.

Dad would wonder in late mindless, asleep, and harmless.

Tiptoe off to bed.

Mom’s undying love.

# Week 17 – Trigonometry

This week in Pre-Calculus 11 we expanded our knowledge on trigonometry. We started off by reviewing Math 10 trig, then moved into unit circles and special triangles. We learned how to do trig without calculators by using special triangles. We reformed SOH CAH TOA to be able to apply it in non-right triangles. Another thing we learned was Sine and Cosine Law.

# DOAS Monologues

The following is an example I created of a monologue for the character Biff from “Death of a Salesman.”

This is a monologue because Biff is talking for a long period of time. This is a monologue because he is talking to Willy.

This monologue would fit into the current plot when Biff, Happy, and Willy are in the restaurant and Willy keeps interrupting Biff. Biff is trying to explain what happened with Oliver

# Week 14 – Multiplying and Dividing Rational Expressions

This week in Pre-Calculus 11 we started the Radical Expressions and Equations unit. In lesson 7.2 we review multiplying and dividing fractions then learned how to divide and multiply rational expressions.

Example 1:

$\frac {24x^3y}{8z^2}\times \frac {2xz^4}{y}$

Before simplifying state the restrictions. Restrictions always come from the denominator.

z $\neq$ 0 and y $\neq$ 0

Simplify

$\frac {48x^4yz}{8yz^2}$

Stop when you can’t simplify any further.

$6x^4z^2$

When multiplying or dividing single variable expressions you need to factor. Factor before stating restrictions because when you factor you can extend trinomials so you have more zeroes. Once the restrictions are stated then you simplify.

How to multiply or divide single variable expressions:

1. Factor

2. State Restrictions

3.Simplify

# Week 12 – Absolute Value Functions

This week in Pre-Calculus 11 we started a new unit. This unit includes absolute values and how we can graph them. I will show you how to graph an abosolute value linear function. There are steps that we need to follow when dissecting an equation.

1. Graph the parent function

2. Find the y-intercept

3. Find the slope

4. Graph absolute value

5. Find the critical point (solution)

Something to look for when graphing linear absolute value functions is the x-intercept. This is important because the x-intercept is the critical point, the point where the graph changes direction. Some equations will have no solution which means they don’t touch the x-axis.

# Week 13 – How to graph reciprocal functions

This week in Pre-Calculus 11 we learned about reciprocal linear and quadratic functions. We learned how to solve the functions algebraically and graphically. My preferred method is graphing to solve because it is a visual for me to see even though it is specific it helps work it out.

How to graph a reciprocal linear function:

1. Graph line (parent function)

2. Plot invariant points (y=1 or y=-1)

3. Asymptotes —> x-intercepts = vertical asymptotes

—> y=0 which is the horizontal asymptotes

How to graph a reciprocal quadratic function:

1. Graph parabola (parent function)

2. Invariant points (y=1 or y=-1)

3.Asymptotes —> x-intercepts = vertical (2 possible solutions)

—> y=0 which is the horizontal asymptote

# Week 11 – Graphing Linear Inequalities in Two Variables

This week in Pre-Calculus 11 we reviewed linear graphing (y=mx+b) from Math 10 and expanded on it. So instead of looking for just one answer of y we look for the possibilities of y. Also we use inequalities instead of an equal sign. I will be blogging about Graphing linear inequalities in two variables (x and y). Solutions to linear inequalities in two variables is represented by the boundary lines and shading on one side.

The inequality used decides what the boundary line looks like.

A solid line is greater/less than or equal to…

A broken line is greater/less than…

When you shade you need to test one of point on either side of the line. If the answer is true such as (y<x+2  —>  0<0+2  —>  0<2) then shade on that side of the line. In the example I used (0,0) as my test point. If it isn’t true choose a point on the other side of the line and test that. Try and use (0,0) as your test point it makes it easier and leaves less room for mistakes.

Examples

This when y is greater than x+2. So the line is broken and the shading is above.

In this graph y is less than x+2. So the line is broken and the shading is on the bottom.

This graph is saying y is greater than or equal to x+2. So the boundary line is solid and the shading is above.

y is less than or equal to x+2 in this graph. So the boundary line is solid and the shading is below.

# Week 10 – Discriminant

This week in Pre-Calculus 11 we reviewed for our midterm. So we are suppose to blog about something we haven’t blogged about yet. I chose the discriminant because it was something I struggled with until review this week. The discriminant is how we determine how many solutions there are. Here is how to find the discriminants $b^2-4ac$. This connects to the quadratic formula, the pink is the discriminant formula.

So you can have 2 solutions, 1 solution, or no solution.

$x^2-6x+5=0$

a=1, b=-6, c=5

$b^2-4ac$

Plug in a, b, and c.

$(-6)^2-4(1)(5)$

36-20

16 gives 2 solutions

This parabola crosses the x-axis twice giving us two solutions.

$x^2-6x+9=0$

a=1, b=-6, c=9

$b^2-4ac$

Plug in a, b, and c.

$-6^2-4(1)(9)$

36-36

0 gives one solution

The parabola crosses the x-axis once meaning there is only one solution.

$x^2-6x+13=0$

a=1, b=-6, c=13

$b^2-4ac$

Plug in a, b, and c.

$-6^2-4(1)(13)$

36-52

-16 gives no solution

The parabola never crosses the x-axis so there is no solution.