You know that it has x+1 as a factor. Completed, the factorization is
The turning point that lies on the x-axis is where it has a root of multiplicity 2.
See which one that is?.
Hello, my cousin has a math problem and needs this problem solved soon. I myself am not sure so any help would be great!
f(x)= x cubed - x squared -5x -3
a) i) Show that (x+1) is a factor of f(x)
ii) Hence or otherwise factorise f(x) fully
b) One of the turning points of the graph of y=f(x) lies on the x-axis. Write down the coordinates of this turning point
It is part b which she needs help with
I typed part a out incase it is needed.
Thanks in advance
A)
i. To show that x+1 is a factor of f(x), you can just use long division and show that it divides without leaving any remainder.
ii. So you can factorize it as (x+1)(quotient). If it needs, you can factorize the quotient too.
B) It says that one of the turning points lies on x axis. First, find the turning points using . You should find two turning points. One of them is on the x axis, plug them in f(x) and see which one is.
Hello, Scotland!
We can use the Remainder/Factor Theorem.
a) i) Show that is a factor of
. .ii) Factorise fully.
. . \text{-}1)^3 - (\text{-}1)^2 - 5(\text{-}1) - 3 \;=\;0" alt="f(\text{-}1) \:=\\text{-}1)^3 - (\text{-}1)^2 - 5(\text{-}1) - 3 \;=\;0" />
Since , then is a factor of
Then we have: .
We can answer this without Calculus . . .b) One of the turning points of the graph of
lies on the x-axis. .Find the coordinates of this turning point.
In part (a), we found that: . x+1)^2(x-3)" alt="f(x) \:=\x+1)^2(x-3)" />
The curve has x-intercepts at .
The intercept has multiplicity 2.
. . Hence, the curve is tangent to the x-axis at
Therefore, there is a turning point at
The graph looks like this:Code:| | * | | | * | --------o----+--------o---- * -1 * | * 3 * * * * | * | * |
Think about the graph y = x+1. The root x = -1 has multiplicity one because it can be written . It crosses the x-axis there because it is negative on one side and positive on the other.
If we have a root of multiplicity 2, we get something like . But is never negative so instead of crossing the x-axis at x = -1, it just touches it, then moves off in the direction it came from. So when we have roots of multiplicity 2, they have a tangent at the axis.
This can be generalised to all even and odd multiplicities.