What is the 3 point curve

Find the parabolic equation from three points

Three points often - not always - define a parabola. This page teaches you how to find the equation and how to determine whether the points actually define a parabola.

Intuition

Move the red dots and observe which values ​​the parameters assume. What if two points have the same $ x $ coordinate? Also lay the points on a straight line.

If two different points have the same $ x $ coordinate, they do not define a function graph: a function is defined, among other things, by the fact that a $ x $ value Not several different $ y $ values ​​may be assigned.

Calculate the function equation

For the following examples I assume that you know the addition and subtraction method for systems of linear equations. If you know the Gaussian method, you can use that too, but I'm not assuming it.

example 1: A parabola goes through the points $ A (-1 | 1) $, $ B (3 | -1) $ and $ C (5 | 7) $. We are looking for a function equation.

solution: If there are three points without special properties such as zeros or the vertex, the general form (polynomial form) $ f (x) = ax ^ 2 + bx + c $ is used.

Each of the three points must "satisfy the equation". For the coordinates of $ A (\ color {# f00} {- 1} | \ color {# 1a1} {1}) $ this means for example
$ \ begin {alignat *} {6} f (\ color {# f00} {- 1}) & \, = \, & \ color {# 1a1} {1} \ quad & a \ cdot (\ color {# f00} {-1}) ^ 2 & \, + \, & b \ cdot (\ color {# f00} {- 1}) & \, + \, & c & \, = \, & \ color {# 1a1} {1} \ \ &&& a & \, - \, & b & \, + \, & c & \, = \, & 1 \ end {alignat *} $

If we do this for all points, we get three equations with three unknowns:

$ \ begin {alignat *} {6} & f (-1) = 1 \ quad && \ text {I} \ quad & a & \, - \, & b & \, + \, & c & \, = \, & 1 \ & f ( 3) = - 1 \ quad && \ text {II} \ quad & 9a & \, + \, & 3b & \, + \, & c & \, = \, & - 1 \ & f (5) = 7 \ quad && \ text { III} \ quad & 25a & \, + \, & 5b & \, + \, & c & \, = \, & 7 \ \ end {alignat *} $

Perhaps you have only solved systems of equations with two unknowns so far. The principle is exactly the same here: the number of unknowns is reduced. Due to the structure, it makes sense to resolve from back to front, i.e. to eliminate $ c $ first. We do this by subtracting Equation I from Equation II:
$ \ text {IV} = \ text {II} - \ text {I} \ quad 8a + 4b = -2 $

This alone doesn't get us anywhere, as there are still two unknowns in this equation. We need another equation that also only contains the unknowns $ a $ and $ b $. To do this, we use Equation III, which we have not yet used, and subtract either I or II:
$ \ text {V} = \ text {III} - \ text {II} \ quad 16a + 2b = 8 $

We can apply the well-known addition method to equations IV and V. In this case it makes no difference whether $ b $ or $ a $ is eliminated as the next element. We choose $ b $ and multiply appropriately before adding:

$ \ begin {alignat *} {6} & \ text {IV} \ quad & 8a & \, + \, & 4b & \, = \, & - 2 \ & \ text {V} \ cdot (-2) \ quad & -32a & \, - \, & 4b & \, = \, & - 16 && \ qquad & | \ text {IV} + \ text {V} \ cdot (-2) \ \ & & -24a & \, \, && \, = \, & - 18 && & | : (- 24) \ & & a & \, \, && \, = \, & 0 {,} 75 \ end {alignat *} $

We can substitute the value for $ a $ in IV or V to determine $ b $:

$ \ begin {align *} a \ text {in IV} && 8 \ cdot 0 {,} 75 + 4b & = - 2 \ && 6 + 4b & = - 2 && | -6 \ && 4b & = - 8 && |: 4 \ && b & = - 2 \ end {align *} $

Now we insert in I, II or III to calculate $ c $:

$ \ begin {align *} a, b \ text {in I} && 0 {,} 75 - (- 2) + c & = 1 \ && 0 {,} 75 + 2 + c & = 1 && | -0 {,} 75-2 \ && c & = - 1 {,} 75 \ end {align *} $

The function we are looking for has the equation $ f (x) = 0 {,} 75x ^ 2-2x-1 {,} 75 $

Example 2: Find the equation of the parabola through the points $ A \ left (- \ tfrac 43 \ big | - \ tfrac 73 \ right) $, $ B \ left (\ tfrac 43 \ big | 3 \ right) $ and $ C (2 | 1) $.

solution: We set up the system of equations again and form the differences of each two equations in order to eliminate $ c $.

$ \ begin {alignat *} {6} & f \ left (- \ tfrac 43 \ right) = - \ tfrac 73 \ quad && \ text {I} \ quad & \ tfrac {16} {9} a & \, - \ , & \ tfrac 43b & \, + \, & c & \, = \, & - \ tfrac 73 \ & f \ left (\ tfrac 43 \ right) = 3 \ quad && \ text {II} \ quad & \ tfrac {16 } {9} a & \, + \, & \ tfrac 43b & \, + \, & c & \, = \, & 3 \ & f (2) = 1 \ quad && \ text {III} \ quad & 4a & \, + \, & 2b & \, + \, & c & \, = \, & 1 \ \ & && \ text {IV} = \ text {II} - \ text {I} \ quad && \, \, & \ tfrac 83b & \, \ , && \, = \, & \ tfrac {16} {3} \ & && \ text {V} = \ text {III} - \ text {II} \ quad & \ tfrac {20} {9} a & \ , + \, & \ tfrac 23b & \, \, && \, = \, & - 2 \ \ end {alignat *} $

In this case, $ a $ has dropped out in equation IV with $ c $. This always happens when the $ x $ coordinates only differ in the sign of two points.
So we can immediately calculate the unknowns:

$ \ begin {align *} & \ text {IV} & \ tfrac 83b & = \ tfrac {16} {3} && |: \ tfrac 83 \ text {or} \ cdot \ tfrac 38 \ && b & = 2 \ & b \ text {in V} & \ tfrac {20} {9} a + \ tfrac 23 \ cdot 2 & = - 2 && | - \ tfrac 43 \ && \ tfrac {20} {9} a & = - \ tfrac {10 } {3} && |: \ tfrac {20} {9} \ && a & = - 1 {,} 5 \ & a, b \ text {in III} & 4 \ cdot (-1 {,} 5) +2 \ cdot 2 + c & = 1 \ && -6 + 4 + c & = 1 && | + 6-4 \ && c & = 3 \ end {align *} $

The equation we are looking for is $ f (x) = - 1 {,} 5x ^ 2 + 2x + 3 $.

Note: If one of the points is the intersection with the $ y $ axis, the parameter $ c $ is already known:
z. B. $ A (0 | 4) \ Rightarrow f (0) = 4 $ $ \ Rightarrow \ underbrace {a \ cdot 0 ^ 2} _ {0} + \ underbrace {b \ cdot 0} _ {0} + c = 4 \ Rightarrow c = 4 $

Then, of course, it doesn't make sense to eliminate $ c $, but instead insert the value immediately and eliminate $ a $ or $ b $. An example of this can be found in the article on Parabola of Two Points and Parameters.

Parable or not?

As you have already noticed in the graphic, three (different) points do not always define a parabola. Obviously this is only then if the abscissas ($ x $ -coordinates) coincide with two points, but not the $ y $ -coordinates: then no function graph is possible.

Example 3: Investigate whether the points $ A (-2 | -2) $, $ B (4 | 3) $ and $ C (16 | 13) $ lie on a parabola or a straight line, and give the corresponding function equation .

Solution 1: We first examine whether the points lie on a common straight line. This is the case if (for example) the slope of the straight line $ (AB) $ coincides with the slope of the straight line $ (AC) $.

$ m_ {AB} = \ dfrac {y_B-y_A} {x_B-x_A} = \ dfrac {3 - (- 2)} {4 - (- 2)} = \ tfrac 56 \ m_ {AC} = \ dfrac {y_C-y_A} {x_C-x_A} = \ dfrac {13 - (- 2)} {16 - (- 2)} = \ tfrac {15} {18} = \ tfrac 56 $

Because of $ m_ {AB} = m_ {AC} $ the points lie on a straight line, and we can use the normal form $ f (x) = mx + n $ (or the point slope form $ f (x) = m (x -x_1) + y_1 $). Any of the three points can be selected to calculate the intercept $ n $, here $ B (4 | 3) $:

$ \ begin {align *} f (x) & = \ tfrac 56 x + n \ 3 & = \ tfrac 56 \ cdot 4 + n && | - \ tfrac {20} {6} \ - \ tfrac 13 & = n \ \ f (x) & = \ tfrac 56x- \ tfrac 13 \ end {align *} $

Solution 2: We do not first check whether the points lie on a straight line, but assume a quadratic function $ f (x) = ax ^ 2 + bx + c $, so proceed as above.

$ \ begin {alignat *} {6} & f (-2) = - 2 \ quad && \ text {I} \ quad & 4a & \, - \, & 2b & \, + \, & c & \, = \, & - 2 \ \ & f (4) = 3 \ quad && \ text {II} \ quad & 16a & \, + \, & 4b & \, + \, & c & \, = \, & 3 \ & f (16) = 13 \ quad && \ text { III} \ quad & 256a & \, + \, & 16b & \, + \, & c & \, = \, & 13 \ \ & && \ text {IV} = \ text {II} - \ text {I} \ quad & 12a & \ , + \, & 6b & \, \, && \, = \, & 5 \ & && \ text {V} = \ text {III} - \ text {II} \ quad & 240a & \, + \, & 12b & \, \, && \, = \, & 10 \ \ & && \ text {IV} \ cdot (-2) \ quad & -24a & \, - \, & 12b & \, \, && \, = \, & - 10 \ & && \ text {V} + \ text {IV} \ cdot (-2) \ quad & 216a & \, \, && \, \, && \, = \, & 0 && \ qquad & |: 216 \ & && & a & \ , \, && \, \, && \, = \, & 0 \ \ end {alignat *} $

We get the other unknowns by inserting:

$ \ begin {align *} & a \ text {in IV} & 12 \ cdot 0 + 6b & = 5 && |: 6 \ && b & = \ tfrac 56 \ & a, b \ text {in I} & 4 \ cdot 0-2 \ cdot \ tfrac 56 + c & = - 2 \ && - \ tfrac 53 + c & = - 2 && | + \ tfrac 53 \ && c & = - \ tfrac 13 \ end {align *} $

You may have suspected that the system of equations is not solvable if the points lie on a straight line. But that is not the case; we get a clear solution. Because of $ a = 0 $, however, the quadratic term is omitted and a linear function is present. As with the first solution, we get the equation $ f (x) = \ tfrac 56x- \ tfrac 13 $.

Which solution is better? There is no general answer to that. If you have to prove that three points do not lie on a parabola, the first way is definitely preferable. If, on the other hand, you should not only check whether three points define a parabola, but also state the equation, the second way is often faster - at the latest when a parabola is available, you have to set up this system of equations. Of course, if your teacher asks you to check the type of function first, you must take the first route.

For interested students and (tutoring) teachers it should be said that the system of equations always has a unique solution if only the $ x $ coordinates are different (keyword Vandermonde determinant).

Exercises

Last update: 02.12.2015; © Ina de Brabandt

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Quadratic Functions 2: Establishing an Equation

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