Writing a Function from its Zeros

If a polynomial function $f\left(x\right)$f(x) of degree $2$2 has the real roots $x=m$x=m and $x=n$x=n, then it must be of the form $f\left(x\right)=a\left(x-m\right)\left(x-n\right)$f(x)=a(x−m)(x−n) with the leading coefficient determinable from extra information given. 

For example, if we knew $x=0$x=0 and $x=8$x=8 were roots of a quadratic function $y=ax^2+bx+c$y=ax2+bx+c, then we would immediately know that the function has the form $y=ax\left(x-8\right)$y=ax(x−8).

If we also knew that the graph of the function passes through $\left(2,36\right)$(2,36) then, upon substitution, we would know that $36=a\left(2\right)\left(2-8\right)$36=a(2)(2−8). Thus $36=-12a$36=−12a and $a=-3$a=−3, so that the function is identified as $y=-3x\left(x-8\right)=-3x^2+24x$y=−3x(x−8)=−3x2+24x. 

Without that extra piece of information we could not identify the specific function. The diagram below shows five other possible candidates that have the same two roots as the identified function.

In our next example, suppose we know that a certain cubic equation has a double root at $x=-1$x=−1 and a single root at $x=10$x=10. Note that the double root implies that the curve of the function has one of two basic shapes as shown here:

We know that the function is of the form $y=a\left(x+1\right)^2\left(x-10\right)$y=a(x+1)2(x−10) and again we are able to determine the leading coefficient when we know another point where the curve passes through.

Suppose we know that the $y$y - intercept is $-5$−5, then by substitution we have that $-5=-10a$−5=−10a and hence $a=\frac{1}{2}$a=12​. This means that $y=\frac{1}{2}\left(x+1\right)^2\left(x-10\right)$y=12​(x+1)2(x−10).

The curve is identified and shown schematically here:

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3