Find the Equation Of the Parabola Containing These Points?

Question

Find the Equation Of the Parabola Containing These Points (1,9) (4,6) and (6,14). Thanks

Answer 1

using standard form ax^2 + bx + c = y, plug in known values for x and y:

a(1^2) + b(1) + c = 9

a(4^2) + b(4) + c = 6

a(6^2) + b(6) + c = 14

a + b + c = 9.....(i)

16a + 4b + c = 6.....(ii)

36a + 6b + c = 14....(iii)

you can solve this using matrices or repeated elimination

ii - i ==> 15a + 3b = -3....(iv)

iii - i ==> 35a + 5b = 5....(v)

3(v) - 5(iv) ==>

105a + 15b = 15

75a + 15b = -15

30a = 30 ==> a = 1

if a = 1, then using (iv) gives 15 + 3b = -3

3b = -18

b = -6

(i) says that a + b + c = 9

1 - 6 + c = 9

c = 14

the equation of the parabola is y = x^2 - 6x + 14

check

1^2 - 6 + 14 = 9 yes (good for (1,9))

16 - 24 + 14 = 6 yes (good for (4,6))

36 - 36 + 14 = 14 yes (good for (6,14))

so this works for the three given points:

y = x^2 - 6x + 14

Answer 2

there are a few ways to do this, try this

realize that any parabola can be written as:

y=ax^2+bx+c  where a, b, c are coefficients

the parabola must go through the points you give above, which means the parabola satisfies the following equations:

9=a+b+c

6=16a+4b+c

14=36a+6b+c

you get these by substituting the pairs of (x,y) for each point

this gives you a 3x3 set of equations which you can solve by techniques you should know from algebra, and get

a=1

b=-6

c=14

or the curve is y=x^2-6x+14