The Video
Video 1 : Factoring Polynomials
To find factors of polynomials is to formed the algebraic long division. For example x-3 is a factor of x3 -7 x - 6?
When dividing x - 3 into x2 - 7 x - 6. First step the problem make a long division problem. There is you dividing x – 3 into x3+ 0 x2- 7x -6. Zero because there is no second degree term. Now, what times x give you x2? Of, course x2, so you multiply x-3 by x2, which give you x3minus 3 x square to get 3 x square. Bring it down next term negative seven x. Dividing x minus 3 into 3 x square minus seven x. Just looking at the first time 3 x square dividing x is 3x. Multiply x minus 3 by 3x. We can get 3 x square minus nine x. Subtracting you have 2x minus 6. Dividing 2 x minus 6 by x-3 which equals 2 and without a remainder. So the solution for a long division problem is x square plus 3 x plus 2. Since x minus 3 divide in x cube minus seven x minus 6. We now know x cube minus 7 x minus 6 equals ( x minus 3) times ( x square plus 3 x plus 2 ). The quadratic expression x square plus 3 plus 2 can be factored into ( x plus 1) times ( x plus 2). So, x cube minus 7 x minus 6 equals ( x minus 3 ) times ( x plus 1) times ( x plus 2 ). Substitution x cube minus 7x minus 6 to zero we get 0=(x-3)(x+1)(x+2) Thus either x-3=0 or x+1=0 or x +2=0 Solving of x we get x=3, x=-1, x=-2. The roots of x cube minus 7 x minus 6 are 3, -1, -2.
Conclusion:
*) 3 roots for this 3rd degree equation
*) 2nd degree equation always have at most 2 roots
*) 4th degree equation would have 4 or fewer roots, and so on.
*) The degree of polynomials equation always limits the number of roots.
Long division process for 3rd order Polynomial:
1. Find a partial quotient of x square, by dividing x into x cube to get x square
2. Multiply x square by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “ clear it out “ or reach a remainder.
Video 2 : Solve The Problems
The next problem is question 13. The figure shows of graph of y equals g of x. If the function h is defined by h(x) = g(2x)+2, what is the value of h(1)? And we are looking for h of 1.
The information of the graph h(x)=g(2x)+2 now we are looking for h(1). We substitute h(1) into this equation. H(1)=g(2)+2 now we get g(2), see when x equals 2, y equals 1. So g(2) is 1. H(1)=g(2)+2, h(1)=1+2, so h(1)=3.
Next is the function with no function problem. This question 13 page 534. Let the function f be defined by F(x)=x+1, if 2f(p)=20, what is the value of f(3p)? We are looking f(3p) what is f when x = 3p?
First information is F(x)=x+1, 2f(p)=20 The figure of 3p start with this equation down here:
2f(p)=20, the, if we are divide by 2 so : f(p)=10. Then f(p) is just what is function f(x) of f(p) f(p)=p+1=10, then p=9. Is this right answer? No, is not. We looking for x=3p, x=27. We have an equation f(x)=x+1, f(27)=27+8=28. The last answer is 28.
Question 17.
In the x y – coordinate plane, the graph of x equals y square minus 4 intersect line l at ( 0,p ) and ( 5,t ). What is the greatest possible value of the slope of l? we’ll be looking for greatest m. The graph intersect in x = 4 line l intersect at ( 0,p ) and ( 5,t ). X is zero, y is p and when x = 5, y=t what is the possible slope for line l. What we are doing now for get the slope? m
equals (y-y1) over (x2-x1) The slope is going to be m=(t-p) over 5. Numerator is t-p we have x=y square minus 4. We can play again the point of intersect ( 0,p ) and ( 5,t ) to the equation x= x square minus 4.
Video 3 : Pre Calculus
Graph of a rational function which can have discontinuities because has polynomial in the denominator.
Is possible value x divide by 0
Example : f(x) equals (x plus 2) over (x minus 1) F(1)= The value become 1+2 over 1-1 equals 3 over zero. That is bad idea.
Graph f(1)=1 plus 2 over zero : Break in finction graph.
F(x) = ( x+ 2) over ( x-1) : insert 0
F(0)=0+2 over 0-1 equals -2. Insert 1 F(1)=1+2 over 1-1 equals 3 over 0 is impossible.
Rational functions don’t always work in this way! Take graph f(x) = 1 over ( x square plus 1 ). Not all rational functions will give zero in denominator because of the +1 ( never zero ).
Rational functions denominator can be zero.
Polynomial have smooth and unbroken curve and for rational function x : zero in the denominator. That impossible situation. A break can show up in two ways. A simply type break is missing point on the graph. Y = ( x square minus x minus 6 ) over ( x minus 3). The graph loose like this if x = 3 ( 3 square minus 3 minus 6 ) over (3 minus 3) equals 0 over 0. That is not possible, not feasible, and not allowed. So that is no way if x = 3. This is a typical example to the missing point syndrome. Y = ((3 square minus 3 minus 6 ) over (3 minus 3) equals 0 over 0. When you see result of 0 over 0 and also tell you direction by possible factor top and bottom of rational function and simplify. For example. Y = ( x square minus x minus 6 ) over ( x minus 3), Equals ( x minus 3 ) times ( x plus 2) over ( x minus 3) so, y = x+2.
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Tampilkan postingan dengan label tugas kuliah. Tampilkan semua postingan
Kamis, 15 Januari 2009
task o(zero)
Tsukuba Journal of Educational Study in Mathematics. Vol.25, 2006
22 1
PURSUING GOOD PRACTICE OF
SECONDARY MATHEMATICS EDUCATION THROUGH
LESSON STUDIES IN INDONESIA
Marsigit
Department of Mathematics Education, Faculty of Mathematics and Science,
the State University of Yogyakarta, Indonesia
Starting in 1999 and lasting in 2005, the extending of IMSTEP_JICA Project
resulting the piloting activities through Lesson Studies for searching good practice of
secondary mathematics teaching in three cluster site West Java, Central Java and
East Java. Results of the studies significantly indicated that there are improvements
Selasa, 30 Desember 2008
SENTENCES
STIPULATION= Ketentuan, syarat; Certainty (rule, direction) that must we do
example :if we want to enter a college, several STIPULATION is waiting us.
ELABORATED= Merinci atau menekuni; an activity(learn,etc) with diligent
example :the student of mathematics should ELABORATED math.
JURING= Section, Segment of the cyrcle
example :in a circle, we can make a SEGMENT, with just one line.
BERSISIAN= Adjacent; from side to side.
example: we can make several adjacent in two form.
example :if we want to enter a college, several STIPULATION is waiting us.
ELABORATED= Merinci atau menekuni; an activity(learn,etc) with diligent
example :the student of mathematics should ELABORATED math.
JURING= Section, Segment of the cyrcle
example :in a circle, we can make a SEGMENT, with just one line.
BERSISIAN= Adjacent; from side to side.
example: we can make several adjacent in two form.
Minggu, 21 Desember 2008
Representing video of learning mathematics (video 3)
Pre-Calculus
• Graps of a rational function
Can have discontinuities
Has a polynomial in the denominator
It’s possible that some value of x we’ll need to division by zero, if so that value is straight up OFF LIMITS
Example: f(x)=(x+2)/(x-1), and take x=1, then we get f(x)=3/0, with zero in the denominator(that’s no good), so choosing x=1 is a bad idea(baaad).
3/0 is break in Function Graph
Not all rational functions will give zero in denominator, for example is f(x)=1/(x2 )+1, the denominator is never zero because of the +1, so the graph is no break, and don’t forget as general rule, when you deal with Rational function, you must expect the possibilities that the denominator is can be zero.
Break can show up in 2 ways:
1. Missing point on the graph,
Example: y=(x2 –x-3)/(x-3), when we take x-3 the result is 0/0, this is Tyical Example of Missing Point Syndrome.
Missing point is a kind of a loophole,
y=(x2 –x-3)/(x-3), when we take x-3 the result is 0/0, but if we simplify first, we got
y=(x2 –x-3)/(x-3)>>>y=(x-3)(x+2)/(x-3)>>>y=x+2.
• Graps of a rational function
Can have discontinuities
Has a polynomial in the denominator
It’s possible that some value of x we’ll need to division by zero, if so that value is straight up OFF LIMITS
Example: f(x)=(x+2)/(x-1), and take x=1, then we get f(x)=3/0, with zero in the denominator(that’s no good), so choosing x=1 is a bad idea(baaad).
3/0 is break in Function Graph
Not all rational functions will give zero in denominator, for example is f(x)=1/(x2 )+1, the denominator is never zero because of the +1, so the graph is no break, and don’t forget as general rule, when you deal with Rational function, you must expect the possibilities that the denominator is can be zero.
Break can show up in 2 ways:
1. Missing point on the graph,
Example: y=(x2 –x-3)/(x-3), when we take x-3 the result is 0/0, this is Tyical Example of Missing Point Syndrome.
Missing point is a kind of a loophole,
y=(x2 –x-3)/(x-3), when we take x-3 the result is 0/0, but if we simplify first, we got
y=(x2 –x-3)/(x-3)>>>y=(x-3)(x+2)/(x-3)>>>y=x+2.
TASK OF ENGLISH
STIPULATION= Ketentuan, syarat; Certainty (rule, direction) that must we do
example :if we want to enter a college, several STIPULATION is waiting us.
ELABORATED= Merinci atau menekuni; an activity(learn,etc) with diligent
example :the student of mathematics should ELABORATED math.
JURING= Section, Segment of the cyrcle
example :in a circle, we can make a SEGMENT, with just one line.
BERSISIAN= Adjacent; from side to side.
example: we can make several adjacent in two form.
From: Bibid Bagasworo
NIM: 07305141013
class: Math R '07
Yogyakarta Math University
example :if we want to enter a college, several STIPULATION is waiting us.
ELABORATED= Merinci atau menekuni; an activity(learn,etc) with diligent
example :the student of mathematics should ELABORATED math.
JURING= Section, Segment of the cyrcle
example :in a circle, we can make a SEGMENT, with just one line.
BERSISIAN= Adjacent; from side to side.
example: we can make several adjacent in two form.
From: Bibid Bagasworo
NIM: 07305141013
class: Math R '07
Yogyakarta Math University
Representing video of learning mathematics(video 2)
We will describe the video that I’ve been looked, the video content some question, these are:
13.the figure shows the graph of y= g(x). if the function h is defined by h(x) = g(2x)+2, what is the value of h(1)?
First piece of information is h(1) then y=g(x) and h(x) = g(2x)+2 then we count h(1)=g(2)+2, refer to the graph g(2)=1 so h(1)=1+2=3 is the answer.
In another question:
13. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
Answer: first information is f(x)=x+1
second information is 2f(p)=20
2f(p)=20f(p)=10, because f(p)=p+1=10, so p equals to 9, we looking for x, so x=3px=27,is this the answer? No, we looking for f(3p)f(27)=27+1=28 is the answer
13.the figure shows the graph of y= g(x). if the function h is defined by h(x) = g(2x)+2, what is the value of h(1)?
First piece of information is h(1) then y=g(x) and h(x) = g(2x)+2 then we count h(1)=g(2)+2, refer to the graph g(2)=1 so h(1)=1+2=3 is the answer.
In another question:
13. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
Answer: first information is f(x)=x+1
second information is 2f(p)=20
2f(p)=20f(p)=10, because f(p)=p+1=10, so p equals to 9, we looking for x, so x=3px=27,is this the answer? No, we looking for f(3p)f(27)=27+1=28 is the answer
Rabu, 17 Desember 2008
Representing video of learning mathematics (video 1)
F(x,y)=0,
Function=f(x) : VLT
Function x=g(y) : HLT :INVERTIBLE
===>y=2x-1 and y=x ;
We can find the intersection by ===>
x=2x-1 ;
1+x=2x ;
1=x ;
so the intersect is 1
2x-1=y
2x=y+1
x=(1/2)(y+1)
x=(1/2)y+(1/2, excange x become y, we get
y=(1/2)x+(1/2)
example;
f(x)=2x+1
g(x)=(1/2)x+(1/2)
f(g(x))=2[(1/2)x+(1/2)]-1
=x+1-1=x, In the other form we have g(f(x))=(1/2)[2x+1]+( 1/2)===>
x-(1/2)+(1/2)=x
g=f-1
f(g(x))=f(f-1)
=x
g(f(x))= f-1(f(x))=x
let’s do one more example, we take y=(x-1)/(x+2),in vertical asymtot in x=-2, and horizontal asymtot in y= 1, the x intercep is going to (1,0), the y intercept is (0,1/2)
we make the equation simple to be calculate:
y(x+2)=x-1
yx+2x=-1-2y
(y-1)x=-1-2y
X=(-1-2y)/(y-1)
Y=(-1-2x)/(x-1)
When x=0 ,y=-1, then when y=0,x=-1/2, so we get graphic that have vertical asymtot x=1,and the horizontal asymtot y=-2,
So f-1 (x)=(-1-2x)/(x-1)
f(f-1 (x))= ([-1-2x/x-1]-1)/([-1-2x/x-1]+2)
=[-1-2x-(x-1)/(x-1)][(x-1)/(-1-2x+2x-2)]
=(-1-2x-x-1)/( -1-2x+2x-2)
=-3x/-3
=x
Back to the front statement, is
f-1 (x)= (-1-2x)/(x-1) and
f(x)=(x-1)/(x+2), so
f-1 (f(x))= (-1-2{(x-1)/(x+2)})/({ (x-1)/(x+2)})-1
=-x-2-2x+2/x-1-x-2
=-3x/-3
=x
F(g(x))=x,wich the range of g is domain of f
G(f(x))=x,wich is the range of f is domain of g
G= f-1 it’s important thought, to remember, the (-1) it’s mean the reciprocal or it’s just mean invers function, when we doubt line we found out the slope of the invers function with the reciprocal of the other slope,that’s true general.
Function=f(x) : VLT
Function x=g(y) : HLT :INVERTIBLE
===>y=2x-1 and y=x ;
We can find the intersection by ===>
x=2x-1 ;
1+x=2x ;
1=x ;
so the intersect is 1
2x-1=y
2x=y+1
x=(1/2)(y+1)
x=(1/2)y+(1/2, excange x become y, we get
y=(1/2)x+(1/2)
example;
f(x)=2x+1
g(x)=(1/2)x+(1/2)
f(g(x))=2[(1/2)x+(1/2)]-1
=x+1-1=x, In the other form we have g(f(x))=(1/2)[2x+1]+( 1/2)===>
x-(1/2)+(1/2)=x
g=f-1
f(g(x))=f(f-1)
=x
g(f(x))= f-1(f(x))=x
let’s do one more example, we take y=(x-1)/(x+2),in vertical asymtot in x=-2, and horizontal asymtot in y= 1, the x intercep is going to (1,0), the y intercept is (0,1/2)
we make the equation simple to be calculate:
y(x+2)=x-1
yx+2x=-1-2y
(y-1)x=-1-2y
X=(-1-2y)/(y-1)
Y=(-1-2x)/(x-1)
When x=0 ,y=-1, then when y=0,x=-1/2, so we get graphic that have vertical asymtot x=1,and the horizontal asymtot y=-2,
So f-1 (x)=(-1-2x)/(x-1)
f(f-1 (x))= ([-1-2x/x-1]-1)/([-1-2x/x-1]+2)
=[-1-2x-(x-1)/(x-1)][(x-1)/(-1-2x+2x-2)]
=(-1-2x-x-1)/( -1-2x+2x-2)
=-3x/-3
=x
Back to the front statement, is
f-1 (x)= (-1-2x)/(x-1) and
f(x)=(x-1)/(x+2), so
f-1 (f(x))= (-1-2{(x-1)/(x+2)})/({ (x-1)/(x+2)})-1
=-x-2-2x+2/x-1-x-2
=-3x/-3
=x
F(g(x))=x,wich the range of g is domain of f
G(f(x))=x,wich is the range of f is domain of g
G= f-1 it’s important thought, to remember, the (-1) it’s mean the reciprocal or it’s just mean invers function, when we doubt line we found out the slope of the invers function with the reciprocal of the other slope,that’s true general.
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