Question 687298
I have the vertex (2,-3) and one point (-1,3)
I need to write a quadratic formula
-----
Form: y = a(x-h)^2+k where (h,k) is the vertex
---
Using the vertex and the point, solve for "a":
3 = a(-1-2)^2-3
6 = a(9)
a = 2/3
-----
Equation:
y = (2/3)(x-2)^2 -3
---
y = (2/3)(x^2-4x+4) - 3
---
y = (2/3)x^2 - (8/3)x - 1/3
==================================
 
also I have the vertex (-1,6) and the y intercept =3 
and I have to write a quadratic formula for that also.
Using the same form:
vertex (-1,6) ; point (0,3)
------
y = a(x-h)^2 + k
3 = a(0--1)^2 + 6
3 = a(1)+6
a = -3
----
Equation:
y = -3(x+1)^2 +6
----
y = -3(x^2+2x+1) + 6
----
y = -3x^2 -6x + 3
=========================
Cheers,
Stan H.
==================