For problems involving quadratics in finance, it is useful to graph the equation. From these, one can easily find critical values of the function by inspection.
Apply the quadratic function to real world financial models
In some financial math problems, several key points on a quadraticfunction are desired, so it can become tedious to calculate each algebraically.
Rather than calculating each key point of a function, one can find these values by inspection of its graph.
Graphs of quadratic functions can be used to find key points in many different relationships, from finance to science and beyond.
The method of graphing a function to determine general properties can be used to solve financial problems.Given the algebraic equation for a quadratic function, one can calculate any point on the function, including critical values like minimum/maximum and x- and y-intercepts.
These calculations can be more tedious than is necessary, however. A graph contains all the above critical points and more, and acts as a clear and concise representation of a function. If one needs to determine several values on a quadratic function, glancing at a graph is quicker than calculating several points.
Consider the function:
Suppose this models a profit function $f(x)$ in dollars that a company earns as a function of $x$ number of products of a given type that are sold, and is valid for values of $x$ greater than or equal to $0$ and less than or equal to $500$.
If a financier wanted to find the number of sales required to break even, the maximum possible loss (and the number of sales required for this loss), and the maximum profit (and the number of sales required for this profit), they could simply reference a graph instead of calculating it out algebraically.
By inspection, we can find that
the break-even points ($x$-intercepts) are between $15$ and $16$ sales, and between $484$ and $485$ sales. The maximum loss before the second break-even points is $750 (the y-intercept), which is lost at $0$ sales. After the second break-even point, the function decreases to infinity, so the losses continue to increase. Maximum profit is $5500 (the vertex), which is achieved at $250$ sales.