How to find a quadratic function from its zeros

In this article we explain how to find a quadratic function if we know its zeros and a point of it, which can be the vertex.

Table of Contents

Deduction of the Procedure

Let *x_1* and *x_2* be two real numbers; we seek to determine a quadratic function *f* that has them as zeros. We turn to the factored form to place the zeros:

*f(x)=a(x-x_1)(x-x_2)*

It is evident that for each value of *a*, there will be a different function that satisfies the condition that its zeros are *x_1* and *x_2*

How to find a quadratic function if we know its zeros — There are infinite quadratic functions that have the same zeros

If, in addition, we know that the function passes through the point *(x_0,y_0),* we can determine a unique function using this data:

*f(x_0)=y_0*

*f(x_0)=a(x_0-x_1)(x_0-x_2)=y_0*

Solve for *a:*

*a=\dfrac{y_0}{(x_0-x_1)(x_0-x_2)}*

This way, we already have the data to find a single quadratic function whose zeros are *x_1* and *x_2* and also passes through the point *(x_0,y_0)*

Steps to find a quadratic function given its zeros

Replace the zero data in the factored form.
Assign different values to the leading coefficient to find quadratic functions with those zeros.
If the function passes through another point, use that information to solve for the unique value of a.
Expand, if necessary, to find the function in polynomial form.

In the case where the given data is that the function has a single zero, it should be interpreted that it has multiplicity 2, using the formula *f(x)=a(x-x_1)^2*

If the given point is the vertex *V=(h,k),* we work with it in the same way as with any other point.

Solved exercises

Exercise 1: Find the quadratic functions whose zeros are *-5* and *1*

Solution:

Replace the zeros in the factored form knowing that *x_1=-5,x_2=1*

*f(x)=a(x-x_1)(x-x_2)*

*f(x)=a(x-(-5))(x-1)*

*f(x)=a(x+5)(x-1)*

For each value of *a*, there will be quadratic functions with zeros at the given numbers. For example:

If *a=1, f(x)=(x+5)(x-1)*

If *a=2, f(x)=2(x+5)(x-1)*

If we expand the products, we get the polynomial form of the function. Doing this in the last case, we have:

*f(x)=2x^2+8x-10*

Exercise 2: Determine the quadratic function whose zeros are *-2* and *14* and passes through the point *(-3,10)*

Solution:

Replace the zeros in the factored form knowing that *x_1=-2,x_2=14*

*f(x)=a(x-x_1)(x-x_2)*

*f(x)=a(x-(-2))(x-14)*

*f(x)=a(x+2)(x-14)*

Now, to determine the value of *a* that makes the function pass through the other point, we use the information that *f(-3)=10*

*f(-3)=a(-3+2)(-3-14)=10*

Solve for *a:*

*a=\dfrac{10}{(-3+2)(-3-14)}=\dfrac{10}{(-1)(-17)}=\dfrac{10}{17}*

So, the requested function is:

*f(x)=\dfrac{10}{17}(x+2)(x-14)*

Exercise 3: Find the quadratic function whose zeros are *0* and *2* and passes through the point *(-1,-5)*

Solution:

Replace the zero data in the factored form: *x_1=0,x_2=2*

*f(x)=a(x-x_1)(x-x_2)*

*f(x)=a(x-0)(x-2)*

*f(x)=ax(x-2)*

Now use the information from the other point, *f(-1)=-5*

*f(-1)=a(-1)(-1-2)=-5*

Solve for *a:*

*a=\dfrac{-5}{(-1)(-1-2)}=\dfrac{-5}{-1(-3)}=\dfrac{-5}{3}*

So, the requested function is:

*f(x)=-\dfrac{5}{3}x(x-2)*

Daniel Machado

Advanced student of Mathematics at Facultad de Ciencias Exactas, Químicas y Naturales. Universidad Nacional de Misiones, Argentina.