Question

f(x) =-X^3 -6x^2 -5x + 12Hi, so basically i have three questions. 1. whatare and how do i find the roots? 2. whats theend behavior? and 3, description of graphwithout a graphing calculator?

Answer 1

Given:

`f\mleft(x\mright)=-x^3-6x^2-5x+12`

Step-by-step explanation:

1. To find: The roots.

By using the remainder theorem,

When x = 1, we get

`f(1)=-1-6-5+12=0`

Therefore, 1 is the root of the polynomial.

That is (x - 1) is one of its factors.

Using the synthetic division,

The given polynomial is reduced to the quadratic,

`\begin{gathered} -x^2-7x-12=-(x^2+7x+12) \\ =-(x^2+4x+3x+12) \\ =-(x(x+4)+3(x+4)) \\ =-(x+4)(x+3) \end{gathered}`

The factored form of the polynomial is,

`f(x)=-(x+4)(x+3)(x-1)`

So, the roots are -4, -3. and 1.

2. To find: The end behaviours.

Since the degree of the polynomial is odd and the leading coefficient is negative.

So, the end behaviours are,

`\begin{gathered} As\text{ }x\rightarrow-\infty,f(x)\rightarrow\infty \\ As\text{ }x\rightarrow\infty,f(x)\rightarrow-\infty \end{gathered}`

3. To describe the graph:

Since all the roots are having odd multiplicity 1.

So, the graph will cross the x-axis at the roots -4, -3, and 1.

Since the degree is 3.

So, the number of turning points is (3 - 1) = 2.

When x = 0, we get y = 12.

So, the y-intercept is (0, 12).