Question

The equation for a parabola has the form y=ax2+bx+c, where a, b, and c are constants and a≠0. find an equation for the parabola that passes through the points (−1,−9), (1,7), and (−6,−14).

Answer 1

The equation of parabola is

`y= ax^2 + bx +c`

For point (-1,-9), equation of parabola is

`-9 = a(-1)^2 +b(-1)+c \\ -9 = a -b+c`

For point (1,7), equation of parabola is

`7 = a(1)^2 + b(1)+c \\ 7 = a + b +c`

For point (-6,-14), equation of parabola is

`-14 = a(-6)^2 +b(-6)+c \\ -14 = 36a -6b +c`

So we have three equations , which are

`a - b + c =-9, \\ a + b + c = 7 \\ 36a -6b+c = -14`

Subtracting first two equation will give

`-2b  = -16 \\ b=8`

Subtracting second and third equation gives

`a+b+c-36a+6b-c = 7+14 \\ -35a +7b = 21 \\ -5a + b =3`

Substituting 8 for b, we will get

`-5a + 8 =3 \\ -5a = -5 \\ a =1`

back substituting 8 for b and 1 for a, we will get

`1+8+c = 7 \\ c = -2`

So we have

`a=1, b=8 and c=-2`

Therefore required equation is

`y= x^2 +8x-2`