# 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.