Equation of a Tangent Line
In short: Equation of a tangent line is y – f(a) = f'(a)(x – a): it describes the straight line that touches a curve at x = a and has exactly the curve’s slope there.
AP Calculus ABCalculusAP Calculus AB Unit 2
What is Equation of a Tangent Line?
The equation of a tangent line describes the one straight line that grazes a curve at a chosen point and heads in the same direction as the curve at that spot. Zoom in far enough and the curve and the line become impossible to tell apart.
Two ingredients are needed: the point \((a, f(a))\) and the slope \(f^{\prime}(a)\). Plugging both into point-slope form gives \(y – f(a) = f^{\prime}(a)\,(x – a)\). The line is a local straight-line model of the function, which is also why it is used to estimate nearby values.
How to spot it
- The problem says “write the equation of the line tangent to the graph at x = …”.
- You are asked to approximate a function value near a known point using a straight line.
- A question asks for the line that touches a curve once and has the same steepness there.
- A normal line is requested, which uses the negative reciprocal of the tangent slope.
The method
- Find the output at the given input: compute \(f(a)\) to get the point \((a, f(a))\).
- Differentiate \(f\) and evaluate \(f^{\prime}(a)\); this number is the slope \(m\).
- Write \(y – f(a) = m(x – a)\), which is the equation of a tangent line in point-slope form.
- Simplify to \(y = mx + b\) if the question wants slope-intercept form, and check that the point satisfies it.
Equation of a Tangent Line: solved example
Problem
Find the equation of the tangent line to \(f(x) = x^3 – 2x\) at \(x = 2\).
Solution
Point: \(f(2) = 8 – 4 = 4\), so the line passes through \((2, 4)\).
Slope: \(f^{\prime}(x) = 3x^2 – 2\), so \(f^{\prime}(2) = 12 – 2 = 10\).
Point-slope: \(y – 4 = 10(x – 2)\), which simplifies to \(y = 10x – 16\). Check: at \(x = 2\), \(10(2) – 16 = 4\).
Try it
Question 1
Write the tangent line to \(y = x^2\) at \(x = -3\).
Show the answer
The point is \((-3, 9)\) and the slope is \(2(-3) = -6\). So \(y – 9 = -6(x + 3)\), or \(y = -6x – 9\).
Question 2
Find the tangent line to \(g(x) = \frac{1}{x}\) at \(x = 2\).
Show the answer
\(g(2) = \frac{1}{2}\) and \(g^{\prime}(x) = -\frac{1}{x^2}\), so \(g^{\prime}(2) = -\frac{1}{4}\). Then \(y – \frac{1}{2} = -\frac{1}{4}(x – 2)\), which is \(y = -\frac{1}{4}x + 1\).
Equation of a Tangent Line vs. Derivative
The derivative at a point is only a number, the slope. The tangent line is a full line with its own equation, built from that number plus the point of contact. Skipping the second ingredient, the output \(f(a)\), is the classic way to lose points.
Keep going
Lessons
Worksheets and quizzes
- The Derivative as a Limit – Worksheet
- The Derivative as a Limit – Quiz
- Linearization and L’Hopital’s Rule – Worksheet
- Linearization and L’Hopital’s Rule – Quiz
Related words
- Derivative
- Limit Definition of the Derivative
- Slope Field
- Implicit Differentiation
- Mean Value Theorem
Whole dictionary: A–Z · AP Calculus AB · Calculus