# Top 5 things I learned in Precalculus 11

Top 5 things I learned in Precalculus 11

Analyzing Quadratic Functions and Inequalities Unit

Key Learning：

Every quadratic function can be written in the form y = ax^2+bx+c, where a, b, and c are real numbers and a is not zero.

1. I learned the properties of quadratic functions and how to analyze the quadratic equation. And I learned that when we are given the quadratic equation in vertex form then we can graph the equation.

Vertex Form $a(x-p)^2 + q = y$

General Form $ax^2 + bx +c = y$

Factored Form $a(x-x_1)(x-x_2) = y$

2. I learned equivalent forms in order to solve equations by using 3 different formulas.

vertex form- $y = a (x-p)^2 + q$, where (p, q) is the vertex of the parabola, and a is the stretch value. It’s the easiest equation to graph because we are given a starting point and the pattern that it goes up by.

general form- $y = ax^2 + bx + c$,  a, b, and c are three real numbers. Once these are given, the values for x and y that make the statement true express a set of (x, y) points which form a parabola when graphed. It’s the most useless equation to graph because we are only given the y-intercept and can’t do much else unless it’s converted to one of the other forms.

factored form– $y = a(x-x1)(x-x2)$,  x1 and x2 are the opposite of the x-intercepts of a graph. This equation is useful only when given the x-intercepts and is used mostly when trying to search for a value as long as there is another point given along the parabola to substitute y and x for to find a.

Rational Expressions And Equations Unit

3. I learned how to simplify rational expressions.

For example, to simplify this expression, we need to factor first. $\frac {(2x+2)}{(x^2+3x+2)}$

Now we can begin with factoring. $\frac{2(x+1)}{(x+2)(x+1)}$

The next step is to look at the denominator and the numerator and cancel out any of them that are identical. This has to be monomial and monomial, binomial and binomial, variable with variable, etc. $\frac {2}{(x+2)}$

Now we can simplify the expression and state the non-permissible values. The non-permissible values are the restrictions and are for the denominator. In this expression, we need to also state the NPVs of the expression before and after we converted it into an all multiplication since it had the division in it.

Therefore x≠-2 or -1.

4. I learned how to solve rational equations.

For example : $\frac{5}{x+4}=\frac {3}{x-3}$

First, you would want to multiply each side by the common denominator $(x+4)$ and $(x-2)$ , which would make the equation become: $(x-2)(5) = (x+4)(3)$

then you would multiply the brackets which would turn the equation into: $5x-10 = 3x+12$

after that, you would just solve from here moving the 3x to one side then moving the 10 too the other side which would have the equation end at:

x = 11

Trigonometry Unit

4. I learned about special triangles.

We use these special triangles when the angle is 30, 60, 45 degrees and also the Multiple of 90 degrees. For these special triangles, we will not have to use a triangle and they give us a more exact answer.

Ex: 5. I learned that Sine law is a way that we can use to find a certain angle or length of something.

For example, for a Triangle ABC from A to B it’s 8.6; from A to C it’s 6.1; Angle C is 65°. Now, we are trying to find the length of BC.

You would use the equation of $\frac{A}{SinA} = \frac{B}{SinB} = \frac{C}{SinC}$ , this is for when you’re looking a missing side but for a missing angle,  and the equation $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$ it’s the same equation just flipped. And what we need to do right now is to find the angle of B, so it can help us find the side length of BC so you would use the equation of $\frac{SinA}{A} = \frac{SinB}{6.1} = \frac{Sin65}{8.6}$ . Now we would use the two fractions that are filled in the most so now the equation would be $\frac{SinB}{6.1} = \frac{Sin65}{8.6}$ .  To solve this equation is to basically isolate the B and put everything else on one side so the equation and solve from there which would give the angle B 40°, and from there you would keep using the same equation to find side BC.

# Week 14 – Math 11

This week in Precal 11 we learned about special triangles. We use these special triangles when the angle is 30, 60, 45 degrees and also the Multiple of 90 degrees. For these special triangles, we will not have to use a triangle and they give us a more exact answer.      # Week 15 – Math 11

For the fourteenth week of Preacul 11, we started a new chapter —Trigonometry.  In grade 11 trig there are way more things to learn and the simplicity of SOH CAH TOA won’t cut it for grade 11 trig.

In this new trig we learn that there are 4 quadrants, it’s different from what we learned in grade 10. In grade 10 we were doing trig in quadrant 1 for most of the time, and we know this because Sine, Cosine and tangent were all positive in quadrant 1. Another thing that we learned was Sine Law, Sine law is a way that we can use to find a certain angle or length of something.

For example, for a Triangle ABC from A to B it’s 8.6; from A to C it’s 6.1; Angle C is 65°. Now, we are trying to find the length of BC.

You would use the equation of $\frac{A}{SinA} = \frac{B}{SinB} = \frac{C}{SinC}$ , this is for when you’re looking a missing side but for a missing angle,  and the equation $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$ it’s the same equation just flipped. And what we need to do right now is to find the angle of B, so it can help us find the side length of BC so you would use the equation of $\frac{SinA}{A} = \frac{SinB}{6.1} = \frac{Sin65}{8.6}$ . Now we would use the two fractions that are filled in the most so now the equation would be $\frac{SinB}{6.1} = \frac{Sin65}{8.6}$ .  To solve this equation is to basically isolate the B and put everything else on one side so the equation and solve from there which would give the angle B 40°, and from there you would keep using the same equation to find side BC.  This is how you use the Sine Law.

# Week 13 – Precalc 11

For the thirteenth week of Precalc 11,  we continued learning about rational expressions and equations.

For example: $\frac{5}{x+4}=\frac {3}{x-3}$

First, you would want to multiply each side by the common denominator $(x+4)$ and $(x-2)$ , which would make the equation become: $(x-2)(5) = (x+4)(3)$

then you would multiply the brackets which would turn the equation into: $5x-10 = 3x+12$

after that, you would just solve from here moving the 3x to one side then moving the 10 too the other side which would have the equation end at:

x = 11

# Week 12 – Precalc 11

For the twelfth week of Precalc 11, we started a new chapter about rational expressions and equations. In this chapter, we have so far covered how to multiply/divide and add/subtract.

To simplify this expression, we need to factor first. $\frac {(2x+2)}{(x^2+3x+2)}$

Now we can begin with factoring. $\frac{2(x+1)}{(x+2)(x+1)}$

The next step is to look at the denominator and the numerator and cancel out any of them that are identical. This has to be monomial and monomial, binomial and binomial, variable with variable, etc. $\frac {2}{(x+2)}$

Now we can simplify the expression and state the non-permissible values. The non-permissible values are the restrictions and are for the denominator. In this expression, we need to also state the NPVs of the expression before and after we converted it into an all multiplication since it had the division in it.

Therefore x≠-2 or -1.

# Week 9 – Precalc 11

For the ninth week of Precalc 11, we learned more about the Analyzing Quadratic Functions and inequalities unit.

Key Learning：

1. Vertex Form $a(x-p)^2 + q = y$
2. General Form $ax^2 + bx +c = y$
3. Factored Form $a(x-x_1)(x-x_2) = y$

We learned equivalent forms in order to solve equations by using 3 different formulas.

vertex form- $y = a (x-p)^2 + q$, where (p, q) is the vertex of the parabola, and a is the stretch value. It’s the easiest equation to graph because we are given a starting point and the pattern that it goes up by.

general form- $y = ax^2 + bx + c$,  a, b, and c are three real numbers. Once these are given, the values for x and y that make the statement true express a set of (x, y) points which form a parabola when graphed. It’s the most useless equation to graph because we are only given the y-intercept and can’t do much else unless it’s converted to one of the other forms.

factored form– $y = a(x-x1)(x-x2)$,  x1 and x2 are the opposite of the x-intercepts of a graph. This equation is useful only when given the x-intercepts and is used mostly when trying to search for a value as long as there is another point given along the parabola to substitute y and x for to find a.

# Week 8 – Precalc 11

For the eighth week of Precalc 11, we started our new unit — Analyzing Quadratic Functions and inequalities.

Key Learning：

Every quadratic function can be written in the form y = ax^2+bx+c, where a, b, and c are real numbers and a is not zero.

First, we learned the properties of quadratic functions and how to analyze the quadratic equation. We learned that when are given the quadratic equation in vertex form then we can graph the equation.

Second, we learned the quadratic equation $ax^2 + bx + c = 0$, this unit we learned $y = a(x-p)^2 + q$. This is combination of the equations $y = (x-p)^2, y = x^2$, and $y = x^2 + q$. These equations are all very important in graphing the curve the equations make.

Example：

Graph y = (x− 2)^2− 4 • Vertex = (2,4)
• x-intercept: 0 and 4
• y-intercept: 0
• opens up (happy)

# Week 4 – Precalc 11

For the fourth week of Precalc 11, we continued studying about “Radical Operation and Equations” unit.

First, we learned how to Multiply & Divide Radicals.

When we are multiplying radicals (with the same index), we need to multiply under the radical first, and then multiply in front of the radical (any values multiplied times the radicals).  When we are dividing radicals (with the same index),  we need to divide under the radical first, and then divide in front of the radical (divide any values multiplied times the radicals).

And also to divide radical expressions with the same index, we use the quotient rule for radicals. If a and b represent nonnegative numbers, where ,  then we have:  # Week 3 – Precalc 11

For the third week of Precalc 11, we started our new unit “Radical Operation and Equations”.

We learned how to write an entire radical to a mixed radical and how to write a mixed radical to an entire radical. And we also learned how to add or subtract radicals. Like terms are radicals with the same index and radicand, these are the only radicals that can be + or- . When there is a variable in the radicand, you must specify what type of number (positive or negative) is needed to make the radical defined, for √ , the radicand must be a positive number.

# Week 2 – Precalc 11

For the second week of Precalc 11, we continued studying about “Roots and Powers”.

We learned power with positive rational exponent and negative rational exponent. We also learned Exponent Laws and Order of Operations which is applying the order of operations and exponent laws to evaluate numerical expressions and simplify algebraic expressions. We can recall “BADMAS” for remembering the order：

B (brackets)  E (exponents)  D/M (divided or multiply )  A/S (add or subtract)

And through the skill check 2, I noticed an exponent law that is very easy to make mistakes, that is !!! $a^0$ = 1 !!!