In order to sketch the graphic of a polinomial we are going to evaluate the points where g(x) = y = 0
[tex]\begin{gathered} 0=-2(3x+4)(x-1)(x-3)^2 \\ 0=(3x+4)(x-1)(x-3)^2 \end{gathered}[/tex]We have that it is satified whenever one or more of the terms of the multiplication is cero, it is to say:
[tex]\begin{gathered} (3x+4)=0 \\ \mleft(x-1\mright)=0 \\ \mleft(x-3\mright?^{})=0 \end{gathered}[/tex]Then, on the following values of x, y=0
[tex]\begin{gathered} x=-\frac{4}{3} \\ x=1 \\ x=3 \end{gathered}[/tex]Then it passes through the coordinates (-4/3, 0), ( 1, 0 ) and ( 3, 0 )
We locate those points in the graphic
Step 2The only way it intercepts x- axis just through those points is if the graph has some curves. We are going to determine how those curves are evaluating the function between those two points
We evaluate it when x = -2, x = 0, x = 2, x = 4
When x = -2
[tex]\begin{gathered} y=-2(3(-2)+4)((-2)-1)((-2)-3)^2 \\ y=-2\cdot(-2)\cdot(-3)\cdot(-5)^2 \\ y=4\cdot(-3)\cdot25 \\ y=-300 \end{gathered}[/tex]When x = 0
[tex]\begin{gathered} y=-2(3(0)+4)((0)-1)((0)-3)^2 \\ y=-2\cdot4\cdot(-1)\cdot9 \\ y=72 \end{gathered}[/tex]When x = 2
[tex]\begin{gathered} y=-2(3(2)+4)((2)-1)((2)-3)^2 \\ y=-2\cdot10\cdot1\cdot(-1)^2 \\ y=-20 \end{gathered}[/tex]When x = 4
[tex]\begin{gathered} y=-2(3(4)+4)((4)-1)((4)-3)^2 \\ y=-2\cdot16\cdot3\cdot1^2 \\ y=-2\cdot48 \\ y=-96 \end{gathered}[/tex]Just as the first step, we know the graph intercepts horizontal axis on the points x=-4/3, x = 1 and x = 3.
We know it intercepts the vertical axis when x = 0. We know, because of Step 2 that when x = 0, then y = 72.
Therefore,
Horizontal intercepts: ( - 4/3, 0), ( 1, 0 ) and ( 3, 0 )
Vertical intercept: (0 , 72)
g has a decreasing behavior on its end
