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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