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13. The figure above shows the graph of y equal g for x. if the function h is defined by h for x equal g for two x plus two. What is the value of h for one?
Problem Solving :
- We are looking for h for one and the first information of the graph listing like it ______
- The next information is
h(x) = g(2x) + 2
- Then substitute 1 to this equation
h(1) = g(2) + 2
- If g(2) = y where is 2 = x, so we can find the value of y by graph. When x = 2 , so we get y = 1. Then we substitute it in the equation
h(1) = 1 + 2
= 3
13. Let the function f be defined by f for x equal x plus one. If two times f for p equals twenty, what is the value of f for three p?
Problem Solving :
- We are looking for f for three p. in another word, “What is f when x equal three p?”
- First information is f for x equal x plus one and the next information is two times f for p equal twenty.
- We start with equation here.
2 f(p) = 20
- Then the function divide by two
F(p) = 10
- What is the p for the situation f for x equal x plus one
f(p) = p + 1 = 10
p = 9
- It is no the answer, because we looking for f when x equal three p
- If we take p equal nine and substitute to x equal three p, so x equal twenty seven. And we have question for function f which is f for x equal x plus one and now we know if
x = 27, so f(27) = 27 + 1
f(27) = 28 (this is the answer)

17. In the xy-coordinate plane, the graph of x = y2 – 4 intersect line l at (0,p) and (s,t). what is the greatest possible value of the slope of l?
Problem Solving :
- We will looking for the greatest slope m.
We can see it peace by peace.
In the xy-coordinate plane the graph of x = y2 + 4 and it is the graph ________
- Whether 2 value that formed by x and x intersect in four.
- Intersect line l in intersection at two points (o,p) and (5,t).
- We write two points that important
______________
And the asked what is the greatest possible value of the slope of l.
- What do we know about the slope will we know that the slope of the line is going to be
Line l = m =
- We applied (0,p) and (5,t) to the line
m =


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FACTORING POLYNOMIALS

- One way defined factors of the polynomial is true of form algebraic long division that look suspiciously like a long division you learned the kids, only harder.
- For example, lets try the shift x – 3 a factor of x3 – 7x – 6
- When dividing x – 3 in to x3 – 7x – 6
First set up the problem make a long division problem for elementary school that is you dividing x – 3 into x3 – 7x – 6

Zero in there is no second degrees term.
- Now you must ask your self what times x give you x cube of course x square. So you write x2 as the polemic question and then multiply x – 3 by x2, which give you x3 – 3x2 which is subtract from x3 + 0x2 to get 3x2.
- Bring it down the next term, make it -7x you have 3x2 – 7x. Now we begin again dividing x – 3 into 3x2 – 7x. Just looking at the first terms, x skill into 3x2 3x times.
- Now 3x is the next part of the answer multiply x – 3 by 3x for applied at all 3x2 – 9x. Subtracting you get 2x – 6. Now we see that x – 3 devise evenly into 2x – 6 which is close 2 without remainder. So the solution to the long division problem x3 - 7x – 6 divide by x – 3 is x2 + 3x + 2.
- Since evenly without reminder then x – 3 is a factor of x3 – 7x – 6
- x2 + 3x + 2 is also a factor of x3 – 7x – 6
- we now know that
x3 – 7x – 6 = (x - 3)( x2 + 3x + 2)
- The quadratic expression (x2 + 3x + 2) can be factored into (x + 1)(x + 2), so
x3 – 7x – 6 = (x - 3) (x + 1)(x + 2)
- Setting the factored formed the equation x3 – 7x – 6 to zero. We get
0 = (x - 3) (x + 1)(x + 2)
Those ever x – 3 = 0
X + 1 = 0
X + 2 = 0
Solving all of this equation for x we get x = 3, x = -1, or x = -2
The roots x3 – 7x – 6 is 3, -1, -2
- Now the three roots for this 3rd degree equation. You remember that the quadratic (2rd degree) equations always have at most 2 roots
- A 4th degree equation would have 4 or fewer roots
- The degree of polynomial equation always limits the number of roots.
- Let summarized long division process for a 3rd order polynomial
1. Find a partial quotient of x2 by dividing x into x3 to get x2
2. Multiply x2 by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “clear it out” or reach a reminder.

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PRE CALCULUS
- Graph of rational function which can have discontinuities has a polynomial in the denominator. Which mean you are dividing by something that is very quantity? So you can be sure what the bottom of the fraction is.
- It is possible that some value of x will meet to division by zero. Example :
f(x) =
and when x = 1 the function value becomes
f(x) =
which is f(x) = , with zero in the denominator.
For this function choosing x = 1 is the bad idea.
- What the bad choice when we make the bottom of the rational function zero. It shows a break in function graph. For example :
Suppose f(x) =
Insert 0 for x, f(x) = = = -2
So, you can put it down on the graph.
- Next to try x = 1, you get f(1) = = , it is impossible.
That means, you can not compute the way value when x = 1.
- Rational functions don’t always work this way.
- Take the graph f(x) =
For this case, what number you choose for x, the denominator never can to be zero. All rational functions will give zero in denominator.
- A break can show up in two ways. The simply type of break is missing point on the graph. For example : y =
- The graph loses like these if x – 3. If you try to substitute x – 3 to equation,
y = = , not allowed.
That is not possible, not feasible, and not allowed.
So, there is no way for x – 3. This is typical example of the missing point syndrome.
- When you see result of and also tell you direction is possible to factor top and bottom of the rational function and simplify.
- Example : y = , the top factor to (x - 3)(x + 2)
y =
the x – 3 on the top cancels with the bottom. So, how function simplifies to
y = x + 2
- Missing point is a loophole. The rational function without simplify, x – 3 is bad point because it means that division by zero. But, if we simplify first, it is no problem for x – 3. That is y = x + 2