Question

Without plotting any points other than intercepts, draw a possible graph of the following polynomial:
f(x) = (x + 8)^3(x + 6)^2(x + 2)(x − 1)3(x − 3)^4(x − 6).

Answer 1

@Michele_Laino

Answer 2

is your function like this:
\[{\left( {x + 8} \right)^3}{\left( {x + 6} \right)^2}\left( {x + 2} \right){\left( {x - 1} \right)^3}{\left( {x - 3} \right)^4}\left( {x - 6} \right)\]

Answer 3

Yes :)

Answer 4

the graph of your function crosses the x-axis at points
x=-8, x=-6, x=-2, x=1, x=3, and x=6
since at those points your function:
\[f\left( x \right) = {\left( {x + 8} \right)^3}{\left( {x + 6} \right)^2}\left( {x + 2} \right){\left( {x - 1} \right)^3}{\left( {x - 3} \right)^4}\left( {x - 6} \right)\]
is equal to zero

Answer 5

now, we have this:
\[f\left( { - 9} \right) = {\left( { - 9 + 8} \right)^3}{\left( { - 9 + 6} \right)^2}\left( { - 9 + 2} \right){\left( { - 9 - 1} \right)^3}{\left( { - 9 - 3} \right)^4}\left( { - 9 - 6} \right) > 0\]

Answer 6

whereas:
\[f\left( 7 \right) = {\left( {7 + 8} \right)^3}{\left( {7 + 6} \right)^2}\left( {7 + 2} \right){\left( {7 - 1} \right)^3}{\left( {7 - 3} \right)^4}\left( {7 - 6} \right) > 0\]

Answer 7

so the graph of your function can be like this: