## anonymous one year ago 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).

1. anonymous

@Michele_Laino

2. Michele_Laino

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

3. anonymous

Yes :)

4. Michele_Laino

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

5. Michele_Laino

|dw:1436971271054:dw|

6. Michele_Laino

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

7. Michele_Laino

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

8. Michele_Laino

so the graph of your function can be like this: |dw:1436971518126:dw|

9. anonymous

Thanks:) You're a LIFESAVER!!!

10. Michele_Laino

thanks!! :)