how to find the equation of a quadratic function given 3 points

In this article, Norman Wildberger explains how to determine the quadratic office that passes through iii points.

Some other application of quadratic functions is to curve fitting, too called the theory of splines. Since a parabola (normalsize{y=ax^2+bx+c}) is specified by 3 numbers, it is reasonable to suppose that we could fit a parabola to 3 points in the plane. This is indeed the instance, and it is a useful idea.

In this step we see how to algebraically fit a parabola to iii points in the Cartesian aeroplane. This involves recalling, or learning, how to solve three equations in three unknowns. This is a useful skill on its own correct.

The unique circle through 3 non-collinear points

A line is adamant past two points. A circle, on the other mitt, is adamant past three points —as long as these points are not collinear (all three points cannot lie on the aforementioned line). The construction of a circle which passes through 3 points is a standard do in Euclidean geometry: we construct the perpendicular bisectors of the line segments determined by these three points, then these 3 lines meet at the circumcenter of the triangle (normalsize{ABC}), namely the centre of the unique circle which passes through all 3 points.

Here is a GeoGebra paradigm of this construction:

pic of the circumcenter of a triangle ABC

What virtually a parabola? Well, if we restrict ourselves to parabolas given by quadratic equations, that is curves which the class (normalsize{y=ax^two+bx+c}), and so since there are three unknown quantities, nosotros can expect that nosotros tin fit such a curve to iii distinct points. Let's come across how this works in practice.

Fitting a parabola to iii points

Suppose we want to discover a parabola with equation (normalsize{y=ax^2+bx+c}) which passes through the points (normalsize{[0,1], [1,5]}) and (normalsize{[2,3]}). Substituting each of the three points into the equation, we get

[Big{one=c}] [Large{5=a+b+c}] [Large{3=4a+2b+c}]

This is three equations in three unknowns. This is more complicated, but in this particular instance, things are simpler since the outset equation already tells us that (normalsize{c=1}), so the other two equations become (normalsize{a+b=iv}) and (normalsize{2a+b=i}).

Solving these equations, yields (normalsize{a=-3}), (normalsize{b=7}) and too (normalsize{c=1}), so the required parabola is (normalsize{y=-3x^ii+7x+1}). Hither is the graph:

Graph of y=-3x^2+7x+1, showing it going through the three points [0,1], [1,5] and [2,3]

Solving three linear equations in three unknowns

Since 2 linear equations represent two lines in the plane, their common solution corresponds to the geometric meet of the 2 lines. For 3 linear equations in iii unknowns, the situation actually corresponds to the mutual intersection point of three planes in three-dimensional infinite!

Fortunately the ancient Chinese were able to develop a general technique for solving such systems of equations. Here nosotros simply attempt to observe a uncomplicated practical method.

Suppose we desire to find the equation of the quadratic office (normalsize{y=ax^2+bx+c}) which passes through the points (normalsize{[ane,3], [2,-ane]}) and (normalsize{[four,1]}). It means nosotros have three equations, one for each of the points – since we know the points given must satisfy the unknown equation. The iii equations are

[Large{3=a+b+c} characterization{b1p} tag {1}] [Large{-1=4a+2b+c} label{b2p} tag {2}] [Large{ane=16a+4b+c}. characterization{b3p} tag 3]

What is the strategy? Information technology is unproblematic: we endeavour to eliminate one of the variables, leaving usa with two equations in ii unknowns. This we know how to solve.

To go two equations in two variables, allow's eliminate (normalsize{c}) from the first two equations. We do that by subtracting i from the other, say decrease the offset from the 2nd:

[Large{-4=3a+b} label{b4p} tag{4}]

Delight make sure you understand how we got that. Now nosotros practise the same with the last two equations: take the third equation and subtract the second:

[Large{2=12a+2b}]

Nosotros can make that a bit simpler by dividing all the coefficients by (normalsize{2}) to get

[Big{1=6a+b}. label{b5p} tag {v}]

At present we treat (normalsize{(ref{b5p})}) and (normalsize{(ref{b4p})}) equally new equations, and utilise them to find (normalsize{a}) and (normalsize{b}).

If we accept (normalsize{(ref{b5p})})–(normalsize{(ref{b4p})}) we get (normalsize{five=3a}), so that (normalsize{a=5/3}), and and so plugging back into either (normalsize{(ref{b3p})}) or (normalsize{(ref{b4p})}) gives (normalsize{b=-nine}). Then putting both of these back into say (normalsize{(ref{b1p})}) gives (normalsize{3=5/3-9+c}) so that (normalsize{c=31/3}).

The parabola is therefore (normalsize{y=five/3x^ii-9x+31/3}). Here is its graph:

Graph of parabola y=5/3x^2-9x+31/3

Q1 (M): Discover the equation of the quadratic function (normalsize{y=ax^ii+bx+c}) which passes through the points (normalsize{[-1,9], [two,3]}) and (normalsize{[5,15]}).

Q2 (C): Find a quadratic relation of the course (normalsize{x=ay^2 + by +c}) which passes through the points (normalsize{[0,1], [1,5]}) and (normalsize{[ii,3]}).

Answers

A1. The answer is the parabolic function (normalsize{y=x^two-3x+5}).

A2. The answer is the curve (ten=- frac{ane}{8}left(3y^2-20y+17right)). Note this is not a function.

Source: https://www.futurelearn.com/info/courses/maths-linear-quadratic-relations/0/steps/12130

Posted by: cooperabelity.blogspot.com