# Finding a quartic equation

1. Mar 3, 2006

### rachael

A quartic function has a equation y=ax^4 + bx^3 + cx^2 + dx + e. Its graph cuts the x-axis at (-1,0) and (2,0). One of these intercepts is a stationary point if inflection. If the graph passes through (1,16), find a, b ,c, d and e.

First i started using y=k(x+a)(x+b)(x+c)(x+d)
then i sub in the x values in the equation
y=k(x-1)(x+2)(X-1)(x+2)
after i sub in (1,16)

is this how i suppose to do it? or it is wrong?

2. Mar 3, 2006

### topsquark

Use the general form: y=ax^4 + bx^3 + cx^2 + dx + e.

You know it passes through (-1,0) and (2,0) and (1,16). That gives you three equations for a,b,c,d,e. Do you recall what a stationary point of inflection point is? (Hint: What points do you look for when you are trying to sketch a graph using Calculus?) Plug in the x-intercepts one at a time and see what the resulting equations can tell you.

-Dan

3. Mar 7, 2006

### rachael

thank you...........

4. Jun 4, 2009

### thinkgreen95

i'm in precalc but am doing a calc ia so could u please explain how to find a quartic given three inflection points only?

5. Jun 4, 2009

### rock.freak667

start a new thread with the relevant information.

6. Jun 7, 2009

### Unit

given that your quartic equation is y = ax^4 + bx^3 + cx^2 + dx + e, to find the inflection points you take the second derivative, so you'd get y'' = 12ax^2 + 6bx + 2c. the roots of this equation show you the inflection points of the original quartic. you can factor the y'' equation (find its two roots). check around with both the original quartic given intercepts, and seeing which one matches the point of inflection.