One of the things you might have noticed is that an nth degree polynomial has at most n-1 "critical" points where the graph levels out (maybe but not necessarily where the graph "turns"). Since this graph has 4 such "critical points" (at x=-3, x= -1. something, x= 0. something, and x= 2) I would try a 5th degree polynomial.
Further, we can see from the graph that the polynomial is 0 at x= -3, x= -1, and x= 2. That means that the polynomial is of the form [tex]y= a(x+3)(x+1)(x- 2)Q(x)[tex] where a is a number and Q(x) is a quadratic polynomial.
To go further, you would have to specify exactly where the turning point ("-1.something", "0.something") and what the y values are there.
So are you saying you think the outline for the equation would be:
Q(x) = x(x+3)(x+1)(X-2)
I don't understand how I would find the 0.something and the 1.something points? How would they relate into the equation?
at x=-3 , x=-1 , x=2 ...... y= f(x)=0
that is 3,1,-2 are its root
such an equation is K (x+3)(x+1)(x-2)=0
as such an equation is zero at x=-3,-1,2
now f(0)=6 or y is 6 when x is 0
put x=0 in the equation
K(0+3)(0+1)(0-2)=6
yielding K=-1
therefore reqd. equation is
f(x)=-(x+3)(x+1)(x-2)
Hello, AceBoogie!
I can get you started . . .
There are three -intercepts . . . two of them are quite special.
At , the curve is tangent to the -axis.
. . Hence, has a factor of
At , the curve passes through the -axis.
. . Hence, has a factor of
At , the curve passes through the -axis.
But there is an inflection point at
. . Hence, has a factor of
Also, has a -intercept at
Can you combine those facts?
Hi,
I strongly encourage you to use your graphing software/calculator to help you with graphing problems. For example, once you have the supposed correct equation, you can check your result. Assuming you can zoom in, zooming in at (-3,0) shows indeed an inflection point. I always use my software to check any graphics problem. Of course, you probably can't do this during a quiz or test. Here's the solution to your problem: