Re: Quadratic Functions...

Hey smyatsallie.

If you have three unknowns, then it means you need three pieces of independent information. Since your polynomial is y(x) = ax^2 + bx + c, it means you need three pieces of information to get your a, b, and c.

You have been given two pieces but you need another. Is there something else that you are assuming? Is the quadratic symmetric around the point x = 0? (If it is then that is your third piece of information)

Re: Quadratic Functions...

Quote:

Originally Posted by

**smyatsallie** Write a quadratic function in vertex form whose graph that has the given characteristics:

a.) vertex at (0,0), passing through the point (6, -9)

The answer is: y= -1/4x^2 But I am not sure how the variable (x^2) fits in there...

I've gotten this far but I believe I am going in the wrong direction...

y=a(x-p)^2+q

-9=a(6-0)^2+0

-9=36a

-1/4=a

So, since you know that p= 0 and q= 0 (you used those above, using the fact that "(p, q)" **is** the vertex), now that you know that a= -1/4, you know that

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

Quote:

b.) vertex (0, -6), passing through the point (3, 21)

If you could help me with these, I would greatly appreciate it. Thank you so much!!

p= 0, q= -6. So 21= a(x- 0)- 6. 21= ax- 6. Solve that equation for a and put it, along with p= 0, q= -6, into y= a(x- p)^2+ q.

(It is the fact that one of the given points is the **vertex** that is the "third piece of information that chiro is talking about.)