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India
Class X

Writing a Function from its Zeroes

Lesson

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(xm)(xn) 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(x8).

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)(28). 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(x8)=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(x10) 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(x10).

The curve is identified and shown schematically here:

 

Worked Examples

QUESTION 1

Form a monic quadratic equation which has solutions $x=0$x=0 and $x=-1$x=1.

QUESTION 2

Determine the equation of a parabola whose $x$x-intercepts are $-10$10 and $4$4, and whose $y$y-intercept is $-40$40.

  1. Express the equation in the following form, for some value of $a$a.

    $y=a\left(x+\editable{}\right)\left(x-\editable{}\right)$y=a(x+)(x)

    Note: you do not need to find the value of $a$a at this point.

  2. Hence determine the value of $a$a.

  3. Hence state the equation of the parabola in the form $y=x^2+\ldots$y=x2+

QUESTION 3

Which of the following could be the equation of the function shown in the graph?

Loading Graph...
A graph of a cubic function on a Cartesian plane , featuring only two x-intercepts at $x=-4$x=4 and $x=1$x=1. The graph touches the x-axis at $x=-4$x=4 and turns around changing its concavity. The graph crosses the x-axis at $x=1$x=1.
  1. $y=\left(x+4\right)\left(x-1\right)^2$y=(x+4)(x1)2

    A

    $y=-\left(x+4\right)\left(x-1\right)^2$y=(x+4)(x1)2

    B

    $y=-\left(x+4\right)^2\left(x-1\right)$y=(x+4)2(x1)

    C

    $y=\left(x+4\right)^2\left(x-1\right)$y=(x+4)2(x1)

    D

Outcomes

10.A.P.1

Zeros of a polynomial. Relationship between zeros and coefficients of a polynomial with particular reference to quadratic polynomials. Statement and simple problems on division algorithm for polynomials with real coefficients.

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