Determine the parabola that passes through the points (1,−2), (2,3), and (3,10)

Answer 1

(1,-2), (2,3), (3,10)

y = (x-h)^2 + k

-2 = (1-h)^2 +k

3 = (2-h)^2 + k

subtract 1st from 2nd to get

5 = (2-h)^2 - (1-h)^2

4-4h+h^2 -(1-2h+h)^2 = 5

4-4h+h^2 -1+2h-h^2 =5

3-2h = 5

2h= -2

h= -1

y=(x-h)^2 + k

y=(x+1)^2 +k

3=(2+1)^2+k

k= 3-9=-6

k=-6

vertex is (-1,-6)

y= (x--1)^2-6

y=(x+1)^2 -6

y = x^2 +2x+1-6

y =x^2 +2x -5

actually the problem is a little more complex

the above calculations assumed a=1

for the upward opening parabola, y=a(x-h)^2 +k

to find a, you'd have 3 unknowns, solved with 3 equations:

-2 =a(1-h)^2 +k

3=a (2-h)^2 +k

10= a(3-h)^2+k

solve by substitution and elimination or matrix algebra

but that all leads to one vertical parabola

there are an infinite number of other parabolas through those 3 points

including a horizontal parabola

x= a(y-h)^2 +k

the three equations are found by substituting the 3 points

1=a(-2-h)^2 +k

2=a(3-h)^2 +k

3 =a(10-h)^2+k

by substitution and elimination or matrix algebra

a=-1/210, h=21.5, k=-762,25'

the leftward opening horizontal parabola through the 3 points is

x=(-1/210)(y-21.5)^2-762.25

but there are also an infinite number of slanted or "tilted" parabolas through the 3 points at different angles

they have the form with xy, x and y terms as well as x^2 or y^2 and a constant term

Still, the easiest, most simple standard parabola is the upward opening parabola, which may be implicityly implied, if the problem isn't more specific Especially if this is an algebra 2 course. The slanted or "tilted" parabola will have a directrix that is not a vertical or horizontal line, but a slanted line, with slope not =0 or undefined.

Answer 2

see solution logic in other answer involving point (5,k)

y=x²+2x-5

check, x=1, y=-3

x=2, y=3

x=3, y=10

Answer 3

Asled and answered.