# Math Help - Writing Equations for Given Polynomial Functions and Parabolas

1. ## 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.

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

2. 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.

3. 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?

4. 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

5. 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:
$(x-2)=0$
$(x-4)=0$
$(x+4)=0$

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

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