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How to Find Equations of Tangent Lines

by Chirantan Basu, Demand Media

    A tangent line touches a curve at one and only one point. The equation of the tangent line can be determined using the slope-intercept or the point-slope method. The slope-intercept equation in algebraic form is y = mx + b, where "m" is the slope of the line and "b" is the y-intercept, which is the point at which the tangent line crosses the y-axis. The point-slope equation in algebraic form is y -- a0 = m(x -- a1), where the slope of the line is "m" and (a0, a1) is a point on the line.

    Items you will need

    • Calculator
    Step 1

    Differentiate the given function, f(x). You can find the derivative using one of several methods, such as the power rule and the product rule. The power rule states that for a power function of the form f(x) = x^n, the derivative function, f'(x), equals nx^(n-1), where n is a real-number constant. For example, the derivative of the function, f(x) = 2x^2 + 4x + 10, is f'(x) = 4x + 4 = 4(x + 1). The product rule states the derivative of the product of two functions, f1(x) and f2(x), is equal to the product of the first function times the derivative of the second plus the product of the second function times the derivative of the first. For example, the derivative of f(x) = x^2(x^2 + 2x) is f'(x) = x^2(2x + 2) + 2x(x^2 + 2x), which simplifies to 4x^3 + 6x^2.

    Step 2

    Find the slope of the tangent line. Note the first-order derivative of an equation at a specified point is the slope of the line. In the function, f(x) = 2x^2 + 4x + 10, if you were asked to find the equation of the tangent line at x = 5, you would start with the slope, m, which is equal to the value of the derivative at x = 5: f'(5) = 4(5 + 1) = 24.

    Step 3

    Get the equation of the tangent line at a particular point using the point-slope method. You can substitute the given value of "x" in the original equation to get "y"; this is point (a0, a1) for the point-slope equation, y - a0 = m(x - a1). In the example, f(5) = 2(5)^2 + 4(5) + 10 = 50 + 20 + 10 = 80. So the point (a0, a1) is (5, 80) in this example. Therefore, the equation becomes y - 5 = 24(x - 80). You can rearrange it and express it in the slope-intercept form: y = 5 + 24(x - 80) = 5 + 24x - 1920 = 24x - 1915.

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    About the Author

    Based in Ottawa, Canada, Chirantan Basu has been writing since 1995. His work has appeared in various publications and he has performed financial editing at a Wall Street firm. Basu holds a Bachelor of Engineering from Memorial University of Newfoundland, a Master of Business Administration from the University of Ottawa and holds the Canadian Investment Manager designation from the Canadian Securities Institute.

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