# Writing Equations for Given Polynomial Functions and Parabolas

• May 31st 2011, 03:33 PM
juvenilepunk
Writing Equations for Given Polynomial Functions and Parabolas
I really have no idea how to do either of those, any help would be very much appreciated. I made the graphs myself so sorry if they're a bit sloppy.

1. ) Write one possible equation for the given polynomial functions.

http://i.imgur.com/sm2h4.png

2.) Write the equation for each parabola in the form f(x)=a(x-h)^2+k

http://i.imgur.com/b4y6N.png
• May 31st 2011, 03:43 PM
Moo
Hello,

1) with the 3 coordinates and the local max & min, that gives you 5 conditions, so you can try a quartic equation ax^4+bx^3+cx^2+dx+e=y
substitute the coordinates, and note that between -4 and 2, the first derivative is 0 and the second is negative. You can choose whatever point you want. Same thing for the point between 2 and 4.

2) substitute the coordinates (-4,0) and (0,2) into a(x-h)^2+k and note that the minimum is attained at -4, so f'(-4)=0 and f''(-4)>0.
• May 31st 2011, 03:53 PM
Quacky
Quote:

Originally Posted by Moo
Hello,

1) with the 3 coordinates and the local max & min, that gives you 5 conditions, so you can try a quartic equation ax^4+bx^3+cx^2+dx+e=y
substitute the coordinates, and note that between -4 and 2, the first derivative is 0 and the second is negative. You can choose whatever point you want. Same thing for the point between 2 and 4.

Alternatively, could it not represent this cubic?
• May 31st 2011, 03:59 PM
juvenilepunk
Quote:

Originally Posted by Quacky
Alternatively, could it not represent this cubic?

The answer my teacher gave us is also on that page,

1.) f(x)= a(x^3-2x^2-16x+32)

2.) y= 1/8(x+4)^2+0
• May 31st 2011, 04:13 PM
Quacky
Quote:

Originally Posted by juvenilepunk
The answer my teacher gave us is also on that page,

1.) f(x)= a(x^3-2x^2-16x+32)

2.) y= 1/8(x+4)^2+0

Do you understand how to get the answers? For 1), I let y=the function.
We know that y=0 when x=2, 4 and -4
So:
\$\displaystyle (x-2)=0\$
\$\displaystyle (x-4)=0\$
\$\displaystyle (x+4)=0\$

Which means that \$\displaystyle y=(x-2)(x-4)(x+4)\$ is a possible solution, because \$\displaystyle y=0\$ for all of the x values provided.

For 2), as Moo said, you've just got to understand the vertex form of a parabola.