Question

How can one graph 3x(x+2)(x+1)? What i did is turn it to 3x^3+9x^2+6x? But now im stuck? Thanks

Answer 1

Interesting question! Write this as y = 3x(x+2)(x+1).
Remember what a "root" or "zero" of a function is? Note that y = 0 for x:{0, -2, -1}.

Answer 2

Sorry forgot to add f(x) at the front. I think you're right that it has to someone turn into y=mx+b

Answer 3

x+2= 0, and x+1= 0. Yes i was able to graph those two as the -2, and -1 plots

Answer 4

Also note that your polynomial begins with 3x^3; in other words, it's a 3rd order polynomial. The coefficient of the first term, 3x^3, is positive, so the graph begins in Quadrant III and moves into Quadrant I as you increase x.

Therefore, we know where the graph of y = 3x(x+2)(x+1) begins and where it ends. We also know that the graph crosses the x-axis (has a root, or has a zero) at 3 different x-values. See my first comment, above.

Answer 5

this is really enuf info to enable you to do a quick sketch of this function. Pls do so.

Answer 6

Where's x=0? Why is x=0 needed?

Answer 7

Refer to the attached plot.

Answer 8

without a calculator/graph how can i determine to get the 0.5, and the 1.5 where the plot bends/curves?

Answer 9

You don't need to. You're asked for just a sketch.
Draw your graph again, but larger in scale. Put dots on the x-axis at -2, -1 and 0.
Now, start drawing a curve in QIII, upward to and through x=-2, then turn downward and draw the curve further through x=-1. Finally, turn the curve upward again and continue drawing it, this time thru x=0. Please do this now.

Answer 10

Your graph should "end" in QI.