You can determine the slope of a tangent line at any point on a function using calculus. The calculus approach requires taking the derivative of the function from which the tangent line originates. By definition, the derivative of a function at any given point is equal to the slope of the tangent at that point. This value is also sometimes described as the instantaneous rate of change of the function. Although calculus has a reputation for being difficult, you can find the derivative to most simple algebraic functions quickly.
This process is sometimes used to find the maximum or minimum values of a curved function, since the tangent line slope will be zero at such points.
Write out the function to which a tangent line is being applied in the form y = f(x). The expression designated f(x) will consist solely of the variable x, possibly occurring several times and raised to various powers, and may also contain numerical constants. As an example, consider the function y = 3x^3 + x^2 - 5.
Take the derivative of the function just written. To take the derivative, first replace every term that is in the form of (a)(x^b) with a term in the form of (a)(b)[x^(b-1)]. If this process results in a term containing x^0, then that x simply takes on a value of "1." Secondly, simply remove any numerical constants. The derivative of the example equation is equal to 9x^2 + 2x.
Determine the x point on the function at which you want to calculate the tangent slope. Insert that value of x into the derivative just calculated and solve for the resulting value of the function. To find the tangent to the example function at x = 3, the value of 9(3^2) + 2(3) would be calculated. This value, 87 in the case of the example, is the slope of the tangent line at that point.
About the Author
Michael Judge has been writing for over a decade and has been published in "The Globe and Mail" (Canada's national newspaper) and the U.K. magazine "New Scientist." He holds a Master of Science from the University of Waterloo. Michael has worked for an aerospace firm where he was in charge of rocket propellant formulation and is now a college instructor.
Ryan McVay/Photodisc/Getty Images