Well you are given the roots.
if we have 3 it would.have to be x^3. So something like:
y = ax^3 + bx^2 + cx + d
this could.also be written:
y = (x + a) (x + b) (x + c)
when you are able to write it like this, we know that the opposite of a, b, and c are roots. this is because if we can make any of the insides of the 3 parenthesis equal 0 then y = 0 and that x.is a root. Well if we know the 3 roots that x will be then we just have to figure out the a, b, and c. So let's plug our roots in.
y = (-1 + a) (-5 + b) (-3 + c)
now we have to make each parenthesis equal 0 to find what a, b, and c should be. It is obvious a = 1 to make.that one zero and b = 5 and c = 3. So we know a, b, and c. now let's plug.those into our first equation.
y = (x + 1) (x + 5) (x + 3)
this is your equation. You can multiply out if necessary
You can write it in two types of fractions:
1) ur answer will be 26 and 2/10 which can be simplified to 26 and 1/5
2) ur answer will be 262/10 which can be simplified to 131/5
The first is the more recommended way of writing your answers.
The function
... y = 1/x
has derivative
... y' = -1/x²
which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.
Your function
... y = (x+1)/(x-3)
can be written as
... y = 1 +4/(x-3)
which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).
The critical point is x=3.
The function is decreasing on (-∞, 3) ∪ (3, ∞), increasing nowhere.
(15/r) - 1...when r = 5
(15/5) - 1 =
3 - 1 =
2 <==
