Given:
[tex]f\mleft(x\mright)=-x^3-6x^2-5x+12[/tex]Explanation:
1. To find: The roots.
By using the remainder theorem,
When x = 1, we get
[tex]f(1)=-1-6-5+12=0[/tex]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,
[tex]\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}[/tex]The factored form of the polynomial is,
[tex]f(x)=-(x+4)(x+3)(x-1)[/tex]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,
[tex]\begin{gathered} As\text{ }x\rightarrow-\infty,f(x)\rightarrow\infty \\ As\text{ }x\rightarrow\infty,f(x)\rightarrow-\infty \end{gathered}[/tex]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).
