Write A Quadratic Function Whose Zeros Are And

Understanding How to Write a Quadratic Function from Its Zeros

The ability to write a quadratic function when given its zeros (also called roots or x-intercepts) is a cornerstone of algebraic manipulation. This skill transforms a graphical concept—where a parabola crosses the x-axis—into a precise algebraic equation. Mastering this process deepens your understanding of the intimate relationship between a quadratic’s solutions and its factored form, empowering you to move seamlessly between different representations of the same function. Whether you're solving real-world problems involving projectile motion or optimizing area, constructing a quadratic from its zeros is an essential tool in your mathematical toolkit.

What Are Zeros, and Why Do They Matter?

A zero of a function is any value of ( x ) for which ( f(x) = 0 ). For a quadratic function ( f(x) = ax^2 + bx + c ), these zeros are the points where the parabola intersects the x-axis. Graphically, they are the solutions to ( ax^2 + bx + c = 0 ). Algebraically, the Factor Theorem provides the critical link: if ( r ) is a zero of a polynomial, then ( (x - r) ) is a factor of that polynomial.

For a quadratic with two distinct zeros, ( r_1 ) and ( r_2 ), the function can be expressed in factored form as: [ f(x) = a(x - r_1)(x - r_2) ] Here, ( a ) is a non-zero real number called the leading coefficient. It controls the parabola’s width and direction (opening up if ( a > 0 ), down if ( a < 0 )). The zeros ( r_1 ) and ( r_2 ) are derived directly from the factors. This form is powerful because it makes the x-intercepts immediately visible without any calculation.

The Step-by-Step Process: From Zeros to Function

To write a quadratic function with given zeros, follow this systematic method:

Identify the Zeros: Clearly note the given zeros. Let’s call them ( r_1 ) and ( r_2 ). They can be integers, fractions, irrational numbers, or even complex numbers (though a real quadratic with real coefficients must have either two real zeros or a complex conjugate pair).
Form the Linear Factors: For each zero ( r ), write the corresponding factor ( (x - r) ). If a zero is positive, the factor becomes ( (x - \text{positive number}) ). If a zero is negative, subtracting a negative yields addition, e.g., a zero of (-3) gives the factor ( (x - (-3)) = (x + 3) ).
Write the Factored Form: Multiply the two linear factors and include the leading coefficient ( a ): [ f(x) = a(x - r_1)(x - r_2) ] If no specific leading coefficient is given, you can use ( a = 1 ) for the simplest form. Sometimes problems specify a particular point the parabola must pass through, which allows you to solve for ( a ).
Expand to Standard Form (Optional but Common): To write the function in the familiar standard form ( f(x) = ax^2 + bx + c ), expand the factored expression:
- Multiply the two binomials: ( (x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) ).
- Then multiply by ( a ): ( f(x) = a[x^2 - (r_1 + r_2)x + (r_1 \cdot r_2)] ).
- Distribute ( a ): ( f(x) = ax^2 - a(r_1 + r_2)x + a(r_1 r_2) ).

This final step reveals the coefficients ( b ) and ( c ) in terms of the zeros and the leading coefficient.

Worked Examples: From Simple to Complex

Example 1: Integer Zeros Write a quadratic function with zeros at ( x = 4 ) and ( x = -2 ). Use ( a = 3 ).

Factors: ( (x - 4) ) and ( (x - (-2)) = (x + 2) ).
Factored Form: ( f(x) = 3(x - 4)(x + 2) ).
Expand: First, ( (x - 4)(x + 2) = x^2 - 2x - 8 ).
Then, ( f(x) = 3(x^2 - 2x - 8) = 3x^2 - 6x - 24 ).
Final Answer (Standard Form): ( f(x) = 3x^2 -

6x - 24).

Example 2: Fractional Zeros Write a quadratic function with zeros at ( x = \frac{1}{2} ) and ( x = -\frac{3}{4} ). Use the simplest leading coefficient ( a = 1 ).

Factors: ( \left(x - \frac{1}{2}\right) ) and ( \left(x - \left(-\frac{3}{4}\right)\right) = \left(x + \frac{3}{4}\right) ).
Factored Form: ( f(x) = \left(x - \frac{1}{2}\right)\left(x + \frac{3}{4}\right) ).
Expand: To avoid messy fractions early, multiply carefully: ( \left(x - \frac{1}{2}\right)\left(x + \frac{3}{4}\right) = x^2 + \frac{3}{4}x - \frac{1}{2}x - \frac{3}{8} = x^2 + \frac{1}{4}x - \frac{3}{8} ).
Final Answer (Standard Form): ( f(x) = x^2 + \frac{1}{4}x - \frac{3}{8} ). (Multiplying through by 8 to clear denominators gives the equivalent integer-coefficient form ( 8x^2 + 2x - 3 ), which is often preferred.)

Example 3: Irrational Zeros and a Specified Point Find the quadratic function with zeros ( x = 2 + \sqrt{5} ) and ( x = 2 - \sqrt{5} ) that passes through the point ( (0, -6) ).

Factors: ( (x - (2 + \sqrt{5})) ) and ( (x - (2 - \sqrt{5})) ).
Factored Form: ( f(x) = a(x - (2 + \sqrt{5}))(x - (2 - \sqrt{5})) ).
Notice the factors are conjugates. Their product simplifies nicely using the difference of squares: ( (x - 2 - \sqrt{5})(x - 2 + \sqrt{5}) = (x-2)^2 - (\sqrt{5})^2 = x^2 - 4x + 4 - 5 = x^2 - 4x - 1 ).
So, ( f(x) = a(x^2 - 4x - 1) ).
Use the point ( (0, -6) ) to find ( a ): ( f(0) = a(0 - 0 - 1) = -a = -6 ) → ( a = 6 ).
Final Answer: ( f(x) = 6(x^2 - 4x - 1) = 6x^2 - 24x - 6 ).

Example 4: A Single Zero (Repeated Root) Sometimes a problem gives only one zero, implying a repeated root (the parabola touches the x-axis at that point). Write a quadratic with a zero at ( x = -1 ) and a leading coefficient ( a = -2 ).

Since the zero is repeated, both factors are the same: ( (x - (-1)) = (x + 1) ).
Factored Form: ( f(x) = -2(x + 1)(x + 1) = -2(x + 1)^2 ).
Expand: ( f(x) = -

2(x^2 + 2x + 1) = -2x^2 - 4x - 2 ).

Final Answer (Standard Form): ( f(x) = -2x^2 - 4x - 2 ).

Conclusion

Finding the equation of a quadratic function from its zeros is a powerful application of the factored form. By understanding that each zero corresponds to a linear factor of the form ( (x - r) ), and that a leading coefficient ( a ) scales the entire function, you can construct the quadratic in factored form and then expand it to standard form. Whether the zeros are integers, fractions, or even irrational numbers, the process remains consistent. In cases where only one zero is given, it often implies a repeated root, leading to a perfect square trinomial. With practice, this method becomes a straightforward tool for building quadratic functions that meet specific criteria, bridging the gap between algebraic expressions and their graphical representations.