how di I find the parabola with these points, (2,65) (4,59)(6,92)(8,79)(10,61)(12,30)?

Answer 1

f(x) = ax^2 + bx + c is an equation of a parabola
Now you know more than three points, so take the first three:
f(2) = 65
f(4) = 59
f(6) = 92
You have three eqns in three unknowns
Solve for a,b,c and you are home in time for ice cream

Answer 2

How do I find the parabola with these points, (2,65); (4,59); (6,92); (8,79); (10,61); (12,30).

Four points will uniquely describe a parabola. You can find the equation of a parabola with three points, but it won't be the only one. If you find the equation of a vertical parabola, in most cases you could also find a horizontal one. So use three points to find a parabola. Use the fourth point to confirm that it has the correct orientation. Let's choose three points.

P(2,65); Q(4,59); R(6,92).

Write three equations and three unknowns assuming that the parabola is vertical.

4a + 2b + c = 65
16a + 4b + c = 59
36a + 6b + c = 92

Solving for a, b, and c we get:

a = 39/8
b = -129/4
c = 110

y = (39/8)x² - (129/4)x + 110
y = 4.875x² - 32.25x + 110

Plug in a fourth point S(8,79) to see if it is also on the parabola. The point S is NOT on the parabola. So this is not a vertical parabola.
_____________

Check to see if it's a horizontal parabola.

P(2,65); Q(4,59); R(6,92).

Write three equations and three unknowns assuming that the parabola is horizontal.

4225a + 65b + c = 2
3481a + 59b + c = 4
8464a + 92b + c = 6

Solving for a, b, and c we get:

a = 13/891
b = -2 127/891
c = 79 553/891

Plug in a fourth point S(8,79) to see if it is also on the parabola. The point S is NOT on the parabola. So this is not a horizontal parabola either.

There are a inifinte number of tilted parabolas thru the three points that do not align with either axis. However, I doubt that there is any parabola that fits thru all the points.

The point R(6,92) in particular looks out of place.

Answer 3

Set up a system of equations; look for Y=ax^2+bx+c.

a4 +b2 +c1 =65
a16 +b4 +c1 =59
a36 +b 6 +c1 =92
a64 +b8 +c1 =79
a100 +b10 +c1 =61

you get the matrix:
4 2 1 65
16 4 1 59
36 6 1 92
64 8 1 79
100 10 1 61

Solve the matrix to get reduced row echelon form. Solving the matrix will give you the coefficients a,b, &c for the parabola passing through those points.