Monthly Archives: October 2018

week 8 -pre-cal 11

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Quadratic Function

the different between Q equation and function are:

the  function could be graph with the y-axis and x-axis,

the equation is to solve to find the root of it and not involve the y variable.

 

6 important points of the Q function are below.

1.Vertex is the point of the maxmium or minimum of the ‘U’ shape graph,

ex: the parents Quadratic function is y=x^2, the vertex is (o,0)

2. line of symmetry is the line in the center of the U shape that each part are symmetric。 the x=o for the parent function. the line of symmetry’s “x” is always the first number of the vertex (0,0)

3. x-intercepte , always has 2 coordinations,

4. y-intercept is when x=0, (0.y)

5, Domain, is the scope of the X of the function,  D of y=x^2 is x=R

6. the Range,is the scope of the Y of the function R of  y=x^2 is y>=0. (the second number of vertex (0,0))

 

The Quadratic function is different from linear function which is a straight line through the entrie graph (ex ,y=X+2), due to the  quadratic function’s highest degree of x is 2 and has 2 roots whereas the linear fiction just has one root and degree of 1, so the shape always shows as a U shape (ex, y=x^2+2).

(linear function doesn’t have a vertex or line of symmetry)

 

 

 

 

week 7 Pre-cale 11

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Factoring the ugly quadratic

this kind of quadratic doesn’t have a method (a+b)^2 or (a-b)^2 , (a+b)(a-b).

Ex1. 0.5x^2 -0.4x-1.2

step 1. changed it to another expression to eliminate the fraction and decimals.

5(x)^2 -4x-12

stept 2. first, we could sure that there are (5x+__)(x+___), through the 5(x)^2 , because the 5 just has two factors are whole number, 1, 5.

step 3. find out the two number which “+ “to -4, and “times” each other to -12, there are must a number is +, and another -. according to the last term, -12.

the factors (possibility) are, 1, -12;   2, -6;   3,-4; the 2 and-6 is the only pair that sum of -4.

step 4. put the 2, and -6 into the half-complete answer,(5x+_2_)(x+_(-6)_), and verified it to adjust the places of 2 and -6 (it easy for people to put them in the wrong places).

step 5,(5x+_6_)(x+_-2__), adjust the initial answer to get the right place and + or “-” or them. because the 5(x)^2 influences the whole question, you need to cautious to think about the 5(x)^2 , because step 3 always just suit the (x)^2 perfectly.

This is my way to solve those kind of questions, it’s not the fastest but the safest and easiest way for me.

 

 

week 6-precal 11 quadratic

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Quadratic

This week we focused on how to solve the quadratic with the smart and quicker ways. there are 2 new ways to solve the Quadratic.

1 , one is factoring the quadratic to two factor x1 and x2 with add a number that make the question could be factored as the perfect square( make the unusual quadratic which without the pattern to become the perfect square in order to further factored.)

Ex  (a+b)^2=(2a)^2+2ab+(b)^2

as a revenue, you know how to factored the quadratic, (2a)^2+2ab+(b)^2 back

but the question give you is look like —

(x)^2+4x=2

then, next step is get the right side be 0, -2 at both side, get

1. (x)^2+4x-2=0

you would stop in this step if you don’t put some stuff in it in order to show a product of  perfect square.

2.  (x)^2+4x+4-4-2=0, put a 4 and a -4, which =0, but it makes much easier.

you can have  3. (x+2)^2-4-2=0

4.  (x+2)^2=6 that +both side+4+2, to leave the perfect square in left side.

5.  \sqrt{(x+2)^2}= \sqrt{6}

6.  (x+2)= +-\sqrt{6},  break the squad root down, *notice the +- signs because there are 2 possibility for the factors, 2 answer for the quadratic with highest degree for 2 , (the#of answers always as same as the highest degree of a quadratic)

7. the x=+-\sqrt{6}-2, (-2 on both side in order to leave the x alone in left)

 

2. another is to use the formula,

The quadratic equation

that works on every quadratic which has the highest degree for 2.

the same question as the ex1. (x)^2+4x=2

1. (x)^2+4x-2=0, the first step is same.

the a = 1 (the coefficient of the x^2), the b =4 ((the coefficient of the x), and the c is the number without the variable which is -2.

then you put it one the formula after you figure out each number is equal which variables. be careful with the +- signs in the front of the square root.

the x always has two answers.

 

week 5 , Percale 11–Factoring polynomial

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factoring polynomial expressions

it the way to revenue the two binomial times each other  that we father break down to one simplest polynomial, we need to find the simplest polynomial back to the factors.

1. Ex. (3x+4)^2 = 9x^2+24 x+16

but in the factoring polynomial, the question would be what the factor is the

12x^2+24 x+16-3^2

step 1, check could the terms be farther simplified  first, so  9x^2+24 x+16

step 2,  check the is the term realted to the commonly methods:

(a+b)^2 , a^2 -b^2 , or (a-b)^2

step 3, 9x^2= (3x)^2, +24 x= (12x \times{2})  +16=(4^2)

so the answer would be  (3x+4)^2

but how to solve a question that couldn’t be factored based on those 3 methods?

2. EX:  5x^2 +9x+4

after make sure that the question couldn’t follow the commonly factory methods, then we could separate the middle term (9x) to two term, 5x and 4 X, in order to have the same factor with 1st and 3rd term ,the (5x)^2, and the 4.

step 1, 5x^2 +5x+4x+4

step 2, find the common factor between the 5x^2 and 5x, and the 4x with 4. treat as two binomials. pick common factor out of the term , then put it in front of  brusket

5x(x+1),   4(x+1)

step 3 get the same factor between the binomials, it one of the factor of the polynomial

the left two terms which  in front of brisket , add them together is the other factor of the original polynomials’ factor.

step 4 ,(x+1) (5x+4) the answer is showed out