Question 926832
your equation is:
f(x) = -3x(x+2)^2 (x-4)^3
(x+2)^2 will have a leading term of x^2
(x-4)^3 will have a leading term of x^3
x * x^2 * x^3 = x^6
your leading exponent is x^6
your leading coefficient will be negative because you have 2 leading coefficients that are positive times 1 leading coefficient that is negative to get a negative leading coefficient because a negative times 2 positives is equal to a negative.
your leading term have an even exponent and your leading coefficient is negative so the equation will fall to the left and fall to the right.
your zeroes are x = 0, x = -2, x = 4
the graph of the equation will touch the x-axis at x = -2 and cross the x-axis at x = 0 and x = 4
this is because the exponent of the (x+2)^2 term is even and the exponent of the (x-4)^3 term and the x term are both odd.
the graph of your equation looks like this:
{{{graph(600,600,-3,5,-1000,1000,-3x*(x+2)^2*(x-4)^3)}}}


here's a reference that can help you understand just what went on.


<a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut35_polyfun.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut35_polyfun.htm</a>