# Thread: Writing An Equation for Polynomial Graph

1. ## Writing 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.

2. ## Re: 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.

3. ## Re: 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?

4. ## Re: 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)

5. ## Re: Writing An Equation for Polynomial Graph

Hello, AceBoogie!

I can get you started . . .

Write an equation for the polynomial graphed in the picture.

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?

6. ## Re: Writing An Equation for Polynomial Graph

Originally Posted by Soroban

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 $\displaystyle (x+3)^3.$
You've encouraged me to finally study the polynomial functions chapter in my book. :P

7. ## Re: 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: