How do you sketch the general shape of #f(x)=-x^3-6x^2-9x-4# using end behavior?

1 Answer
Sep 4, 2017

Answer:

You need both the solutions and the end behavior to sketch this graph well. There are two solutions at #x=-1# and one at #x=-4#. The right end of the graph should go down, and the left should go up. graph{-x^3-6x^2-9x-4 [-10, 10, -5, 5]}.

Explanation:

Because the highest variable (aka the "degree" of the function) is odd, the ends of the graph go in opposite directions.
Think about the cubic parent function, where the right side goes up; the left side goes down; and there's a little wiggle at the origin. It's the same idea with all odd functions: the ends go in opposite directions, and there are wiggles in between.

Now, since the #x^3# term is negative, the directions are reversed. The right side should go down and the left up. That's a big part of the task of sketching the graph, but you need some solutions too:

I used the rational roots test to find that #-1# is a solution, so I know that #(x+1)# is a factor.

I used synthetic division to factor it further, and I got #-x^3-6x^2-9x-4= -(x+1)(x^2+5x+4)#
Factoring the quadratic gives you #-(x+1)(x^2+5x+4)=-(x+1)(x+1)(x+4)#, so your solutions are at #x=-1,-1,-4#.

Since we have two solutions at #x=-1#, we know there's a "bounce" there, like the bounce at #x=0# in your quadratic parent function #y=x^2#.

Give this information, you can sketch the part of the graph going up and left from #x=-4#. There's a bounce at #x=-1#, and it's easy to add a little more specificity to the graph by finding where the y-intercept is.
The graph passes through the y axis where #x=0#, so plug #x=0# into the function to find the #y#-intercept. All the terms with x end up#=0#, and all you have left is #f(0)=-4#. So you have another point for your sketch at #(0,-4)#, and you're good to go!