Constructing quadratics

It is a fundamental fact that two points uniquely determine a line. That is, given two distinct points, there is one and only one line that passes through these points.

How many (distinct) non-collinear points in the plane are required to determine a parabola? The answer is three.

This idea may be expressed by the following theorem:

Theorem

If two quadratic functions \(f(x) = ax^2+bx+c\) and \(g(x)= Ax^2+Bx+C\) take the same value for three different values of \(x\), then \(f(x)=g(x)\) for all values of \(x\).

Proof

We start by observing that a quadratic equation can have at most two solutions. Hence, if the quadratic equation \(dx^2+ex+f=0\) has more than two solutions, then \(d=e=f=0\).

Now, if \(f(x)\) and \(g(x)\) agree for \(x=\alpha\), \(x=\beta\) and \(x=\gamma\), then

\[ f(x)-g(x) = (a-A)x^2+(b-B)x+(c-C)=0 \]

when \(x= \alpha\), \(x=\beta\) and \(x=\gamma\). In that case, the quadratic equation above has three different solutions and so the coefficients are 0. Thus, \(a=A\), \(b=B\) and \(c=C\), and so \(f(x)=g(x)\) as claimed. \(\Box\)

Example

Find the equation of the quadratic function whose graph passes through \((-1, 2)\), \((0,3)\) and \((1,6)\).

Solution

Suppose the equation is \(y=ax^2+bx+c\). Substituting the coordinates of the three points, we have

\[ 2=a-b+c, \qquad 3=c, \qquad 6=a+b+c. \]

Thus, \(c=3\) and so \(a-b=-1\) and \(a+b=3\). Therefore \(a=1\) and \(b=2\).

Hence the function is \(y=x^2+2x+3\).

Exercise 11

Express \(x^2\) in the form \(a(x-1)^2+b(x-2)^2+c(x-3)^2\).

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