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Introduction to Derivatives · Ages 17–18

Finding the Equation of a Tangent Line

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Intended for 12th-grade calculus students. Use derivatives to determine slopes, identify points of contact, and write tangent line equations in slope-intercept form.

  1. Find the equation of the tangent line to f(x) = x^2 at x = 3.

  2. Find the equation of the tangent line to f(x) = 3x^2 - 5x + 2 at x = 2.

  3. Find the equation of the tangent line to f(x) = x^3 - 2x at x = -1.

  4. Find the point on the graph of f(x) = x^2 + 6x where the tangent line has a slope of 4.

  5. Find the equation of the tangent to f(x) = 2x^2 that is parallel to the line y = 8x - 3.

  6. Find the equations of all horizontal tangent lines to the curve f(x) = x^3 - 3x.

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Finding the Equation of a Tangent Line

Answer key

  1. Find the equation of the tangent line to f(x) = x^2 at x = 3.

    Point: (3, 9). Slope: f'(3) = 2(3) = 6. Equation: y = 6x - 9.

  2. Find the equation of the tangent line to f(x) = 3x^2 - 5x + 2 at x = 2.

    Point: (2, 4). Slope: f'(x) = 6x - 5, so f'(2) = 7. Equation: y = 7x - 10.

  3. Find the equation of the tangent line to f(x) = x^3 - 2x at x = -1.

    Point: (-1, 1). Slope: f'(x) = 3x^2 - 2, so f'(-1) = 1. Equation: y = x + 2.

  4. Find the point on the graph of f(x) = x^2 + 6x where the tangent line has a slope of 4.

    Set f'(x) = 2x + 6 = 4, yielding x = -1. Substitute into f(x) to get y = -5. Point: (-1, -5).

  5. Find the equation of the tangent to f(x) = 2x^2 that is parallel to the line y = 8x - 3.

    Parallel slope is 8. Set f'(x) = 4x = 8, so x = 2 and y = f(2) = 8. Equation: y - 8 = 8(x - 2), or y = 8x - 8.

  6. Find the equations of all horizontal tangent lines to the curve f(x) = x^3 - 3x.

    Set f'(x) = 3x^2 - 3 = 0, giving x = 1 and x = -1. At x = 1, y = -2 (line: y = -2). At x = -1, y = 2 (line: y = 2).