how do u state restrictions on a variable?

Question

iwas absent when they did this n we hav test tomoro so i need
to know how to do it

state any restrictions on the variable, then simplify

x^2-x-6
-------------
x^2+3x+2

Answer 1

I'm only guessing here but I'd say the restrictions your talking about are the asymptotes. This is when something is divided by a variable in such a way that makes the number equal to infinity.

i.e.
if you had 100/x as x approaches 0 (gets smaller) the value of 100/x will approach infinity and hence does not have an answer.

In your case you need to factor x^2 + 3x + 2 to find this value which will give you your asymptotes.


x^2 + 3x + 2 = 0
(x+2) (x+1) = 0
therefore x cannot equal to -2 or -1 as this would give you division by zero, something that cannot occur.



/edit
woops forgot to simplify,

(x+2)(x-3)
= --------------
(x+2)(x+1)

(x-3)
= ------
(x+1)

Answer 2

After factoring the 1st bit, you get (x^2 -sixteen / x^2 +5x+6) =(x - 4)(x + 4) / (x + 3)(x + 2) regulations take place once you have 0 as a denominator (what's a million/0? undefined) (x +3)(x +2) have been interior the denominator so it can not equivalent 0 x +3 = 0 x = -3 x + 2 = 0 x = -2 so some distance, we've 2 regulations, -3 and -2 the subsequent bit is (x^2 + 5x+4 / x^2 - 2x -8) = (x + a million)(x + 4) / (x - 4)(x + 2) via fact (x - 4)(x + 2) is a denominator, it can not = 0 x - 4 = 0 x = 4 x + 2 = 0 x = -2 to this factor, we've x can not = -3, -2 and four there is greater... (x^2 -sixteen / x^2 +5x+6) / (x^2 + 5x+4 / x^2 - 2x -8) = (x^2 -sixteen / x^2 +5x+6) cases (x^2 - 2x -8) / (x^2 + 5x+4) ......................the reciprocal we've got a sparkling denominator (x^2 + 5x + 4) = (x + a million)(x + 4) x + a million = 0 x = -a million x + 4 = 0 x = -4 all of the regulations are x can not = to -3, -2, -4, -a million, 4

Answer 3

You don't want to have division by zero, and the problem states to state restrictions before simplifying, so
x^2 + 3x + 2 ≠ 0
Factoring you find that
x^2 + 3x + 2 = (x + 1)(x + 2)
If either factor is 0, the expression = 0, so your restriction is refined to x ≠ { -1, -2}.
Now, factoring the numerator you find that
x^2 - x - 6 = (x + 2)(x - 3),
giving you
(x + 2)(x - 3)
---------------- =
(x + 1)(x + 2)

(x - 3)
--------
(x + 1)

Note that the simplified expression has a value when
x = -2, but the original expression does not. That's why you must state your restrictions before simplifying.

Answer 4

restrictions are numbers that x cannot be in order to make the function undefined, in other words,....any number that can make the function undefined is considered a restriction, so simply the only thing that can restrict a fraction is if the denominator is 0
*Restriction is also called domain of a function btw*
so factor the bottom
x^2+3x+2 factors into --> (x+2)(x+1)
equate both factors to zero...
and your x = -1, -2
and that is your domain or restriction, x cannot be -1, or -2 which will make the function undefined, you do not need to worry about the top because there are no restrictions to the top, it can turn into anything and still make the function acceptable, even if it is 0 or a negative number, changing the top will not make the function undefined

Answer 5

The restrictions on problems of this form (a number divided by another number) is to examine when you are dividing by zero. Since, dividing by zero is not defined, you must restrict values of x that make the divisor zero.

In your case x^2+3x+2 must not be allowed to be zero.

Factor it: (x+2)(x+1) --> therefore x must not equal -2 or -1

for it it were allowed, then you would be dividing by zero.