Given three points, find the equation of a parabola that passes through the points...?

Given: axis of symmetry is parallel to the x-axis

Means: general equation is x = Ay² + By + C

Given: A (7, 5)

Means: x = 7, y = 5

Plug these into the equation.

x = Ay² + By + C

7 = A(5)² + B(5) + C

7 = A(25) + 5B + C

7 = 25A + 5B + C

Given: B (-1, 3)

Means: x = -1, y = 3

Plug these into the equation.

x = Ay² + By + C

-1 = A(3)² + B(3) + C

-1 = A(9) + 3B + C

-1 = 9A + 3B + C

Given: B (17, 0)

Means: x = 17, y = 0

Plug these into the equation.

x = Ay² + By + C

17 = A(0)² + B(0) + C

17 = A(0) + 0 + C

17 = 0 + C

17 = C

Your system of equations is:

25A + 5B + C = 7

9A + 3B + C = -1

0A + 0B + C = 17

For systems with 3 or more variables, I prefer using matrices and Cramer's rule.

The matrix for D is:

......[25 5 1]

D = [9 3 1]

......[0 0 1]

Repeat the first two columns and evaluate for D.

......[25 5 1 25 5]

D = [9 3 1 9 3]

......[0 0 1 0 0]

D = (25)(3)(1) + (5)(1)(0) + (1)(9)(0) - (0)(3)(1) - (0)(1)(25) - (1)(9)(5) = 75 + 0 + 0 - 0 - 0 - 45 = 30

The matrix for DA is:

........[7 5 1]

DA = [-1 3 1]

........[17 0 1]

Repeat the first two columns and evaluate for DA.

........[7 5 1 7 5]

DA = [-1 3 1 -1 3]

........[17 0 1 17 0]

DA = (7)(3)(1) + (5)(1)(17) + (1)(-1)(0) - (17)(3)(1) - (0)(1)(7) - (1)(-1)(5) = 21 + 85 + 0 - 51 - 0 - -5 = 60

The matrix for DB is:

........[25 7 1]

DB = [9 -1 1]

........[0 17 1]

Repeat the first two columns and evaluate for DB.

........[25 7 1 25 7]

DB = [9 -1 1 9 -1]

........[0 17 1 0 17]

DB = (25)(-1)(1) + (7)(1)(0) + (1)(9)(17) - (0)(-1)(1) - (17)(1)(25) - (1)(9)(7) = -25 + 0 + 153 - 0 - 425 - 63 = -360

The matrix for DC is:

........[25 5 7]

DC = [9 3 -1]

........[0 0 17]

Repeat the first two columns and evaluate for DC.

........[25 5 7 25 5]

DC = [9 3 -1 9 3]

........[0 0 17 0 0]

DC = (25)(3)(17) + (5)(-1)(0) + (7)(9)(0) - (0)(3)(7) - (0)(-1)(25) - (17)(9)(5) = 1275 + 0 + 0 - 0 - 0 - 765 = 510

Evaluate for A, B, and C.

A = DA / D = 60 / 30 = 2

B = DB / D = -360 / 30 = -12

C = DC / D = 510 / 30 = 17

Update the general equation.

x = Ay² + By + C

x = 2y² - 12y + 17

ANSWER: x = 2y² - 12y + 17

CHECK:

x = 2y² - 12y + 17 with A (7, 5)

7 = 2(5)² - 12(5) + 17?

7 = 2(25) - 60 + 17?

7 = 50 - 43?

7 = 7?

true

x = 2y² - 12y + 17 with B (-1, 3)

-2 = 2(3)² - 12(3) + 17?

-2 = 2(9) - 36 + 17?

-1 = 18 - 19?

-1 = -1?

true

x = 2y² - 12y + 17 with C (17, 0)

17 = 2(0)² - 12(0) + 17?

17 = 2(0) - 0 + 17?

17 = 0 + 17?

17 = 17?

true

Since the axis of symmetry of the parabola is parallel to the x-axis, the equation of the parabola will be

x = Ay^2 + By + C

Substituting the three points on the parabola (7, 5) (-1, 3) (17, 0), create 3 linear equations as below

Eqn1 25A + 5B + C = 7

Eqn2 9A + 3B + C = -1

Eqn3 0A + 0B + C = 17

From these 3 equations solve for A, B and C

A = 9/5, B = -57/5 and C = 17

The equation of the parabola will be

(9/5)y^2 - (57/5)y + 17 = 0

Let f(x) = ax^2 + bx + c be your parabola.

You know three points, so:

f(7) = 5

f(-1) = 3

f(17) = 0

Plug in each of those points into the generic equation for f(x). For example, for the first one, you will get:

f(x) = ax^2 + bx + c

f(7) = a*7^2 + b*7 + c

f(7) = 49a + 7b + c

And because f(7) = 5. you know that:

5 = 49a + 7b + c

You have one equation in terms of a, b, and c. Find the other two equations with the other two points. After you do that, you will have three equatoins, and three unknowns. Pick any two equations, and eliminate any variable you like. Then pick a different pair of equations, and eliminate the same variable. Then take THOSE two equations in two variables, and solve via substitution. Finally, go back and get the third variable from any one of the original equations.

Finally, check your answer. Write down f(x), and plug in a, b, and c, and see that your three points work. It's not a quick problem, particularly if they don't give you either both zeros or at least give you a y-intercept.

If it is parallel to the X axis , it takes a form (y-k)^2 =4p(x-h) , (h,k) vertex or ,

y^2 +Ay +Bx + C =0 then plug points, isolate A, B, C

25+5A+7B+C=0 or 5A+7B+C=-25

9+3A -B + C=0 or 3A-B+C = -9

17B +C =0 or 17B+C=0

Solve for A, B, C .-