How to Derive a Quadratic Function from Given Conditions

In this article, we will go through the process of finding a quadratic function given specific conditions. We'll start by using the vertex form of a quadratic equation and then convert it to the standard form. This is a fundamental skill in understanding and manipulating quadratic functions, which are widely used in various applications, from physics to economics.

Understanding the Vertex Form and its Relation to the Standard Form

The vertex form of a quadratic function is given by:

$f(x) a(x-h)^2 k$

Where $(h, k)$ is the vertex of the parabola. The standard form, on the other hand, is:

$f(x) ax^2 bx c$

The relationship between these forms is crucial for analyzing the properties of the quadratic function. Let's explore this using a specific example.

Solving for the Quadratic Function

Consider a quadratic function of the form $f(x) ax^2 bx c$ that has a vertex at $(2, -5)$ and passes through the point $(3, 1)$. Our goal is to determine the values of $a, b,$ and $c$ and express the function in both the vertex form and the standard form.

Using the Vertex Form

Given that the vertex is at $(2, -5)$, we can write:

$f(x) a(x-2)^2 - 5$

We know that the function passes through the point $(3, 1)$. Substituting $x 3$ and $f(3) 1$, we get:

$1 a(3-2)^2 - 5$

Simplifying the equation:

$1 a - 5$

$a 6$

Substituting $a 6$ back into the vertex form:

$f(x) 6(x-2)^2 - 5$

Expanding this to the standard form:

$f(x) 6(x^2 - 4x 4) - 5$

$f(x) 6x^2 - 24x 24 - 5$

$f(x) 6x^2 - 24x 19$

Deriving from the First Derivative

Alternatively, we can use the first derivative approach. The first derivative of the quadratic function is given by:

$f'(x) 2ax b$

The function has a vertex at $(2, -5)$, so the first derivative vanishes at $x 2$.

$0 2a(2) b$

$b -4a$

Using the points we have, we get the system of equations:

$4a(2)^2 b(2) c -5$

$9a(3)^2 b(3) c 1$

$2a b 0$

Simplifying these equations:

$16a 2b c -5$

$27a 3b c 1$

Substituting $b -4a$ into the first two equations:

$16a 2(-4a) c -5$

$32a - 8a 19 -5$

$-4a c -5$

$27a 3(-4a) c 1$

$27a - 12a c 1$

$15a 19 1$

Solving the system:

$-4a c -5$

$15a 19 1$

$a 6, b -24, c 19$

Therefore, the function in standard form is:

$f(x) 6x^2 - 24x 19$

Conclusion

We have successfully derived the standard form of the quadratic function given specific conditions. This method involves using the vertex form, which is derived from the known vertex and a point the function passes through, and the algebraic manipulation to find the coefficients. Understanding these concepts is essential for more advanced applications in mathematics and real-world problem-solving.