Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,8).

y=-(x+3)(x+1)(x-2)=-x^3-2x^2+5x+6

Step-by-step explanation:

If the zeros are:

x1=-3

x2=-1

x3=2

then the function is:

y = a (x-x1) (x-x2) (x-x3) = a (x+3) (x+1) (x-2)

y = a (x^3+2x^2-5x-6)

To find the value for "a" we use the information that, if x=1, then y=8:

y = a (1+2-5-6) = a (-8) = 8 therefore a=-1

So:

y=-(x+3)(x+1)(x-2)=-x^3-2x^2+5x+6

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)