find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1)

Question

Pre-calculus, getting the general equation of a parabola

Answer 1

Problem: find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1)
Pre-calculus, getting the general equation of a parabola

We are looking for an equation in the form of y = ax^2 + bx + c.
Given three points, we can replace x and y with real values to
create three equations in three unknowns: a, b and c.

a(-1)^2 + b(-1) + c = -3
a(1)^2 + b(1) + c = -2
a(2)^2 + b(2) + c = 1

a(1) + b(-1) + c = -3
a(1) + b(1) + c = -2
a(4) + b(2) + c = 1

1) a - b + c = -3
2) a + b + c = -2
3) 4a + 2b + c = 1

We can use that system of equations to find the values
for what are actually the constants in the general equation.

Subtract equation 2 from equation 1.

   a - b + c = -3
-(a + b + c = -2)
---------------------
     -2b      = -1
4) -2b = -1
b = -1/-2 = 1/2          <<<<<<<<<<<<<<

Subtract equation 3 from equation 2.

     a +   b + c = -2
-(4a + 2b + c =  1)
------------------------
 -3a -    b      = -3
5) -3a - b = -3

Multiply equation 5 by 2.

2(-3a - b) = -3 * 2
6) -6a -2b = -6

Subtract equation 6 from equation 4.

         -2b = -1
-(-6a -2b = -6)
-------------------
   6a        =  5
6a = 5
a = 5/6          <<<<<<<<<<<<<<

Use equation 3 to solve for c.

4a + 2b + c = 1
4(5/6) + 2(1/2) + c = 1
10/3 + 1 + c = 1
10/3 + 1 + c - 1 = 1 - 1
10/3 + c = 0
10/3 + c - 10/3 = 0 = 10/3
c = -10/3
c = -3 1/3          <<<<<<<<<<<<<<

The equation is y = 5/6 x^2 + 1/2 x - 3 1/3

 

Answer 2

find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1)

The equation of a parabola with a vertical axis is:

(y – k) = a.(x – h)^2

The three points coincident with the curve are,

(-1, -3), (1, -2), (2, 1).

These coordinate points satisfy the equation of the parabola.

Substituting for these coordinate values into the eqn of the parabola gives three simultaneous equations.

(-3 – k) = a.(-1 – h)^2

(-2 – k) = a.(1 – h)^2

(1 – k) = a.(2 – h)^2

Expanding the squared terms,

-3 – k = a(1 + 2h + h^2)   ------------- (1)

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

1 – k  = a(4 – 4h + h^2)   ------------- (3)

   (1) - (2) gives

-1 = 4ah -> ah = -1/4

   (2) - (3) gives

-3 = a(-3 + 2h)

-3 = -3a + 2ah

-3 = -3a – 2/4   using ah = -1/4.

3a = 2 ½ = 5/2

a = 5/6

Therefore h = -1/(4a) = -1/(20/6) = -6/20 = -3/10

h = -3/10

substituting for a = 5/6 and h = -3/10 into (1 – k) = a.(2 – h)^2,

(1 – k) = 5/6.(2 + 3/10)^2

1 – k = 5/6.(23/10)^2

1 – k = 5*23^2/(6*100) = 2645/600 = 529/120

k = 1 – 289/120 = -409/120

k = -409/120

 

The unknown parameters are: h = -3/10, k = -409/120, a = 5/6

The parabola is: (y + 409/120) = 5/6.(x + 3/10)^2, or

The equation of the parabola is: y(x) = (5/6)x^2 + (1/2)x – 10/3

 

Comments

Fermat, I have plotted your equation and found that it does indeed hit the given points. However, I read the problem as specifying a vertical parabola, rather than a horizontal one. What is it that gives you the impression that the parabola is horizontal? Did I miss something?

CWA

---

You're absolutely right. I misread it.

Apologies to the poster. I'll fix it now.