Hey all,

Hoping someone can help me figure out how to write an equation for the polynomial graphed in the picture. Really lost, if any help could be given it would be appreciated.

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- Jun 17th 2013, 04:44 PMAceBoogieWriting An Equation for Polynomial Graph
Hey all,

Hoping someone can help me figure out how to write an equation for the polynomial graphed in the picture. Really lost, if any help could be given it would be appreciated.

Attachment 28638 - Jun 17th 2013, 05:02 PMHallsofIvyRe: Writing An Equation for Polynomial Graph
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. - Jun 17th 2013, 05:13 PMAceBoogieRe: Writing An Equation for Polynomial Graph
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? - Jun 18th 2013, 02:57 AMmpx86Re: Writing An Equation for Polynomial Graph
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) - Jun 19th 2013, 07:11 PMSorobanRe: Writing An Equation for Polynomial Graph
Hello, AceBoogie!

I can get you started . . .

Quote:

There are three $\displaystyle x$-intercepts . . . two of them are quite special.

At $\displaystyle x=2$, the curve is tangent to the $\displaystyle x$-axis.

. . Hence, $\displaystyle f(x)$ has a factor of $\displaystyle (x-2)^2.$

At $\displaystyle x = \text{-}1$, the curve passes through the $\displaystyle x$-axis.

. . Hence, $\displaystyle f(x)$ has a factor of $\displaystyle (x+1).$

At $\displaystyle x = \text{-}3$, the curve passes through the $\displaystyle x$-axis.

But there is an inflection point at $\displaystyle (\text{-}3,0).$

. . Hence, $\displaystyle f(x)$ has a factor of $\displaystyle (x+3)^3.$

Also, $\displaystyle f(x)$ has a $\displaystyle y$-intercept at $\displaystyle (0,6).$

Can you combine those facts?

- Jun 19th 2013, 11:23 PMReneGRe: Writing An Equation for Polynomial Graph
- Jun 28th 2013, 07:38 PMjohngRe: Writing An Equation for Polynomial Graph
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:

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