Math 11 – Yi's Blog

Week 17- Preclac 11

yil2016 on January 14, 2019

Sin Law : $\frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}$ or $\frac{Sin A}{a}=\frac{Sin B}{b}=\frac{Sin C}{c}$

We only can use sin law when we know the value of one angle and the length of it’s opposite side and extra value of angle or length.

Ex: a=5, b= 2, angle A=30. Find the value of angle B

$\frac{Sin 30}{5}=\frac{Sin B}{2}=Sin 30\cdot2=Sin B\cdot5$ ;

$B=\frac{1}{5}$

Week 16 -Precalc 11

yil2016 on December 24, 2018January 6, 2019

A rotational angle is found between the initial arm and terminal arm ( A positive angle is a counter clockwise rotation, a negative angle results from a clockwise rotation )

Standard position is when the initial arm is on the positive x-axis rotation is about the origin.

Ex: show angle 220 in standard position

Rotational angle: 220

Reference angle: 40

Coterminal angle: -140

If the reference angle = A, rotational angles of this reference angle can be expressed by A, 180-A, 180+A, 360-A.

In addition, we can use three letters to show trigonometric function.

( Ex: tan A = $\frac{y}{x}$ , sin A $\frac{y}{r}$ , cos A $\frac{x}{r}$ )

Ex: Find all of the rotational angles of the reference angle 27, the range of rotational angle is 0~360.

1) rotational angle = A = 27

2) rotational angle = 180-A = 153

3) rotational angle = 180+A = 207

4) rotational angle = 360-A = 333

Week 15- Precalc 11

yil2016 on December 16, 2018

The most important point I learned in this week was solving rational equations.

Situation 1: Each side of equal sign only one fraction. (In this situation, we can simply use cross multiply.)

Ex: $\frac{x-2}{5}\cdot\frac{x-1}{3}$

$3x-6=5x-5$ , $-1=2x$ , $x=\frac{-1}{2}$

Situation 2: One side of equal sign have more than one fractions.

Step 1: Eliminate all fractions. (each fractions multiply the smallest common denominator)

Step 2: Solve the simplified equation.

Step 3: Get the result.(include the non-permissible value of x)

Ex: $\frac{5}{x+3}\cdot\frac{2}{x^2+9}=\frac{4}{x-3}$

$5\cdot(x-3)\cdot2=4\cdot(x+3)$ , $5x-4x=12+13$ , $x=25$

Week 14 – Precalc 11

yil2016 on December 10, 2018February 14, 2019

In this week, I learned how to multiplying and dividing rational expressions.

To multiply, first find the greatest common factors of the numerator and denominator. Next, regroup the factors to make fractions equivalent to one. Then,multiply any remaining factors.
To divide, first rewrite the division as multiplication by the reciprocal of the denominator.

Ex:

Week 13 – Precalc 11

yil2016 on November 30, 2018December 2, 2018

In this week, I learned how to graph the absolute value of quadratic equations. I concluded that we only need 3 steps to graph it:

-graph parent function

-find x-intercepts (critical point)

-graph (find all points below x-axis reflect them up)

Also I learned how to graph the reciprocals of quadratic functions.

-graph parent function

-find invariant points (y=1 or y=-1)

-find asymptotes (x=x-intercept and y=0)

-graph hyperbola

Ex: graph the absolute value of quadratic function $y=2x^2-2$

graph the reciprocals of quadratic function $y=2x^2-2$

In addition, the reciprocals of quadratic function have three extinct type

Week 12- Precalc 11

yil2016 on November 25, 2018November 25, 2018

In this week, I learned how to graph the absolute value equations and how to solve it. Assume that there have an absolute value equation: y=|x-2|. The first thing I will going to do is graph the equation without the absolute value bar. (draw the graph of y=x-2 ).

We need make sure the graph of y=|x-2| above the x-axis because the f(x) of absolute value must greater or equal to zero. Reflect the negative part of this graph form the x-intercept, so that we can draw the graph of y=|x-2|.

Solving the absolute value equations:

Step1 : Isolate the absolute value equations.( Make sure let two equations equal to positive and negative the quantity on the other side of the equation. )

Step2 : Solve the equations

Step3 : Check the solutions

Ex: 25=|-2x+9|

(1)25=-2x+9 and -25=-2x+9

(2)25-9=-2x and -25-9=-2x

16=-2x and -34=-2x

x=-8 and x=17

(3)when x=-8, 25=|-2(-8)+9|=|16+9|=|25|

when x=17, 25=|-2(17)+9|=|-34+9|=|-25|

x=-8 and x=17 are the solutions of this absolute value equations.

Week 11-Precalc 11

yil2016 on November 18, 2018

In class of pre-calculus 11 this week we learned about graphing quadratic inequalities. I’ve simply concluded that using two steps to draw an inequality with two variables. Firstly, draw the parabola and the parabola’s function is to isolate the regions. Take $y-5x^2>-3$ for example, we should draw the parabola of $y-5x^2=-3$ to isolate the whole rectangular plane coordinate system. Secondly, shadow the area by finding a test point wherever and substitute into the inequality. If I using point(0,0), the inequality will become to $0-5(0)^2>-3$ . The inequality is correct it means that the area of point(0,0) exists should be shadowed.

Ex: Graph each quadratic inequalities.

(1) $y\geq -2x^2+8$

(2) $y<-1.5x^2+6$

Week 10- Precalc11

yil2016 on November 12, 2018November 12, 2018

We finished the midterm text and started a new Chapter 5-Graphing Inequalities and Systems of Equations. In first class of this chapter we learned how to solve the quadratic inequation in one variable. Firstly, there have four circumstances:

(1) $ax^2+bx+c>0$

(2) $ax^2+bx+c<0$

(3) $ax^2+bx+c\geq0$

(4) $ax^2+bx+c\leq0$

We can use five steps to solve each quadratic inequation.

Step 1: Factor the inequation.

(Ex $x^2-9x+14>0=(x-2)(x-7)>0$ )

Step 2: Find the value of x which can lead the equation in left part become to zero.

(Ex $x_{1}=2,x_{2}=7$ )

Step 3: Draw a number axis and label the value x in the same line.

Step 4: Find three text point in each area to find the correct area.

(When x=0, y>0; x=5, y<0; x=10, y>0)

Step 5: Conclusion.

(The domain of this quadratic inequation is x<2 and x>7)

In addition, draw the circle open when we represent x on the number axis in the first and second circumstances and draw the circle filled in the other situations.

Week 9- Precalc11

yil2016 on November 5, 2018November 12, 2018

This week, we mainly reviewed the mid-term exam.So I don’t learn much new knowledge. I think the most important thing I learned this week is how to use vertex to solve practical problems.

Ex: Find two integers with a sum of 24 and the greatest possible product.

(1) Let a=first number and b=second number

(2) We can get two distinct equations : a+b=24 and ab=Product

(3) Transform a+b=24 to a=24-b and substitute into the second equation. We can get a new equation: (24-b)b=Product

(4) Find the solution of this equation by completing prefect square.

(24-b)b=Product

$24b-b^2=Product$ ,

$-(b^2-24b+144-144)=Product$ ,

$-(b-12)^2+144=Product$

(5) We get a coordinate (12,144),it means (b,maximum product)

(6) Calculate the value of a by substitute value of b into original function.

(7) Conclusion: the two integers are 12 and 12

Week 8-Precalc 11

yil2016 on October 28, 2018October 28, 2018

In this week, we learned how to transform the standard form of quadratic equation to vertex form.(The major reason why we should learn how to transform it to vertex form is that we get the coordinate of vertex directly) Easily, we only need to find the value of a,q,p in the formula $y=a(x-p)^2+q$ . If there have a standard form $y=ax^2+bx+c$ and we can translate it to vertex form by complete perfect square. Consider the $y=x^2+4x+7$ as an example.