Representing the video of learning mathematics

13. The figure above 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) ?
We are looking for h(1)
The function is h(x) = g(2x) + 2
loking for h(1), substitute 1 to the equation, we get h(1) = g(2) + 2
While in the figure, shows that value y in g(2) is 1.
So, if we substitute g(2) as 1, we get h(1) = 1+2
= 3

13. Let the function f be defined by f(x) = x+1
If 2 times f(p) = 20, what is the value of f(3p) ?
f(3p) what is f when x = 3p?
In here we have two equations, there are f(x) =x+1 and 2f(p) = 20
We can dividing the equation 2f(p) = 20 by 2. So we get f(p) = 10.
And then assuming x equal p, from the equation f(x) = x+1 we get equation f(p) = p+1.
Then we can substitute value of f(p) to equation f(p) = p+1
f(p) = p+1
10 = p+1
so we get p = 9
we are looking for f when x = 3p, so we get x = 27
and we get f(27) = 27 + 1 = 28.
17. in the xy-coordinate plane, the graph of x = y2 - 4
Intersect line l at coordinate (0,p) and (5,t) . What is the greatest possible value of the slope of l ?
We will looking for greatest solpe (m)
x = y2 – 4 intersect line l
x y
0 p
5 t
We know te slope of line l is m = y_(2- y_1 )/x_(2- x_1 )
We have the value of (x1,y1) = (0,p) and (x2,y2) = (5,t)
So, the slope of the line l is m = (t-p)/5

Video 2
Factoring polynomials
When we the find factor polynomial is true follow Algebraic long division
For example x-3 is factor of x^3-7x-6
We can get that statement with operate x^3-7x-6 divide by x-3
First, select the problem long division problem there is x^3+0x^2-7x-6
Then operate x^3 divide x. we get x^2. Then multiply x^2 with x-3 so we get x^3-3x^2. Subtract x^3+0x^2-7x-6 with x^3-3x^2. Ringing down the therm 7x we get 3x^2-7x. And then we look first therm , 3x^2 divide x we get 3x. multiply 3x with x-3, we get 3x^2-9x. 3x^2-7z subtract by 3x^2-9x we get 2x. Ringing down the therm 6 we get 2x-6. 2x-6 devide x we get 2, then multiply 2 with x-3 we get 2x-6. Then 2x-6 subtract 2x-6 no remainder. The solution of the long division problem x3-7x-6 devide by x-3 is x^2+3x+2.
Since x^3-7x-6 devide by x-3 is no remainder, so x-3 is a factor of x^3-7x-7. The cloution wich is x^2+3x+2 is also a factor of x^3-7x-6. We now know x^3-7x-6 equals x-s times x^2+3x+2 . and x^2+3x+2 can be factor into x+1 times x+2. So x^3-7x-6 equals x-3 times x+1 times x+2. Sading the factor for the equation x^3-7x-6 into zero, we get 0 equals x-3 times x+1 times x+2. Then x-3 equal 0, or x+1 equal 0, or x+2 equal 0. Solving all the equation for x we get x equal 3, x equal -1, and x equal -2.
So, we can conclude that the roots of x^3-7x-6 are 3, -1, and -2.
3 roots for 3th degree equation, quadratic (2th egree) equation always have at most 2 roots. A 4th degree equation would have 4or fewer roots
The degree of polynomial equation always limit the number of roots
Long division for a 3rd order polynomial :
Find a partial quotient of x^2, by dividing x into x3 to get x2.
Multiply x^2 by the divisor & subtract the product from the dividend.
Repeat the process until you either “clear it out” or reach a remainder.

Video 3
Pre – Calculus
Graph of a rational function
Which can have discontinuities because the graph of a rational function has a polynomial in the denominator.
Is possible same valuer x devide by 0
Example :
f(x) = (x+2)/(x-1)
when x=1 the function value become (1+2)/(1-1)
we get f(1) = 3/0 it is a bad idea.
Graph f(1) = (1+2)/0 is break in function graph
f(x) = (x+2)/(x-1) insert 0 become f(x) = (0+2)/(0-1) = -2
f(x) = (x+2)/(x-1) insert 1 become f(x) = (1+2)/(1-1) = 3/0 is imposible
rational function don’t always work this way
take graph f(x) = 1/(x^2+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 andf 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
example :
y = (x^2-x-6)/(x-3) in the graph look like break
if x = 3 we get (3^2-3-6)/(3-3) = 0/0 that is not possible, not feasible, not allowed
so that is no way for x = 3, this is a typical example to the missing point syndrom
y = (3^2-3-6)/(3-3) = 0/0
when you see result of 0/0 and also tell you direction be possible.
Factor top and bottom of rational function and simplify y
for example : y = (x^2-x-6)/(x-3)
= ((x-3)(x+2))/(x-3)
= x + 2




Video 4
Inverse function
F(x,y) = 0
Function y = f(x) is Vertical Line Test
Function x = g(y) is Horizontal Line Test : invertible
For example:
y = x2
x = g(y) : Horizontal Line : Invertible
if we draw the graph of function y = x2
from the 1 function so in the graph intersect horizontal line in two points.
y = 2x -1
if we draw graph of function y = 2x -1 in coordinate xy in points ( 0,-1 )
and ( 1/( 2),0 ) then draw graph of the function y = x
from the two functions,
substitute y = x to y = 2x-1
x = 2x-1
x + 1 = 2x
1 = x
From the function 2x -1 = y we get 2x = y + 1
x = 1/2 (y + 1)
x = 1/2 y + 1/2

change x become y so we get new function y = 1/2 x + 1/2
draw graph of y = 1/2 x + 1/2 through ( 0, -1) and ( 0, 1/2 )
so invers line y = 2x -1 in line y = x is y = 1/2 x + 1/2

from all that we calculate above, we get:
f(x) = 2x – 1
g(x) =

f( g(x) ) we substitute g(x) to variable on the function f.
g( f(x) ) we substitute f(x) to variable on the function g.
f ( g(x) ) = 2 ( g(x) ) - 1
f ( g(x) ) = 2( - 1
f ( g(x) ) = x + 1 – 1
f ( g(x) ) = x

g ( f(x) ) =
g ( f(x) ) = )
g ( f(x) ) = x
g =
f ( g(x) ) = f ( (x) ) = x
g( f(x) ) = f ( (x) ) = x


Method 2 find y^-1

y ( x+2 ) = x - 1
xy + 2y = x – 1
xy – x = -1 -2y
x ( y – 1 ) = -1 -2y
x = (-1-2y)/(y-1)
so y^-1 = (-1-2x)/(x-1)