SOLUTION: I need help with sketching the graph of this factored polynomial {{{y=(x+1)^4 (x-2)^3 (x-1)}}}

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Question 369474: I need help with sketching the graph of this factored polynomial
y=%28x%2B1%29%5E4+%28x-2%29%5E3+%28x-1%29

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
I need help with sketching the graph of this factored polynomial 
y=%28x%2B1%29%5E4+%28x-2%29%5E3+%28x-1%29

The roots, zeros, x-intercepts, or solutions 

(they are called by all four names at different times and are the
places where the curve comes in contact with the x-axis) 

are found by setting the expressions in the parentheses = 0

x+1=0, x-2=0, x-1=0

and solving, getting

x=-1, x=2, and x=1

When you have something like this:  %28x-R%29%5EN as a factor
of a polynomial then we say R is a zero with multiplicity of N.  
If N is an even number the graphed curve "bounces" off the x-axis 
at R, and if N  is an odd number, the graphed curve cuts through 
the x-axis at R.

Since the exponent 4 in %28x%2B1%29%5E4 is even, we know that the 
graphed curve "bounces off" the x-axis at -1.

Since the exponent 3 in %28x-2%29%5E3 is odd, we know that the 
graphed curve cuts through the x-axis at 2.

Since the understood exponent 1 in %28x-1%29 is odd, we know that the 
graphed curve cuts through the x-axis at 1.

The y-intercept is found by substituting 0 for x in

y=%280%2B1%29%5E4+%280-2%29%5E3+%280-1%29

y=%281%29%5E4%28-2%29%5E3+%28-1%29

y=%281%29%28-8%29%28-1%29

y=8

So we draw the graph this way.  We start at the y-intercept and draw
downward to the left so that the curve bounces off the x-axis at -1.
Then to the right it eventually goes down and cuts through the x-axis
at 1 and 2. 

Here is the graph

graph%284000%2F17%2C800%2C-2%2C3%2C-5%2C12%2C%28x%2B1%29%5E4+%28x-2%29%5E3+%28x-1%29%29

The graph cuts through the x-axis at 1 and 2 and "bounces off" the 
x-axis at -1.  To draw it accurately like this you would
have to plot some more points.

Edwin