Step-by-step explanation:
This a polynomial.
The graph leading degree is 3 so as of right now we know it will have a S shape.
First, we know that graphs will have at most two infliction points, which means the graph changes curvature two times at most.
So when graphing polynomials there are serial steps.
Step 1: Finding the zeroes of the function.
The constant term is 6 and the leading coeffecient is 1 so the possible roots (x values of this function that will make it equal zero.)
(1,-1,2,-2,-3,3,6,-6).
So let see which values makes this function equal 0.
-1,2, and -3 works so this means our roots are
[tex](x + 1)(x + 3)(x - 2)[/tex]
So we know our graphs will have points (-1,0) (-3,0) and (2,0).
Tip: Apply Synethic Division in the future for higher degree polynomials after finding roots.
Step 2: See how the function behaves around the zeroes.
So let see between the zeroes , -3 and -1. First, plugging point close to -3. into the function and plot those.
Let try -2. Plugging in -2 we get ,4 so plot (-2,4). Let try -1.5, plugging that in we get 2.625. We should see the function increasing temporarily before we plug in -1.5. So plug in point, we get
(-1.5,2.625)., so plot that.. Eventually plug points that are close to x=-1, and plot them.. So it seems as after we reached the function local max, we start to decreasing until we reach our zero.
We also have a y intercept 0,-6 so plot that point. Using g that info, after plugging points between zeroes -1,2 we get see the function decreasing but start to change shift to increasing around when x=0.8. Then our function passes through last zero.
Final Step :End behavior
Since this function has a odd degree and positive leading coeffecient,
as x approaches infinity, f(x) approaches infinity
as x approaches negative infinity, f(x) approaches negative infinity.
So draw a line going upwards after passing through x=2, and a line going downwards from x=-3.
Our graph finally should look like this. See picture on answer.
Describe the graph: A S shaped with zeroes at x=-3,-1, and 2.
Solutions are -3,-1, and 2.
When h(x)=0, the graphs x values are -3,-1,and 2.
