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MathematicsHigh schoolMedium

Equation of the tangent line to a graph

Find the equation of the tangent line to the graph of the function

f(x)=x3−2xf(x) = x^3 - 2x

at the point with xx-coordinate x0=1x_0 = 1.

Hintshow

The tangent line at x0x_0 has the equation y=f′(x0)(x−x0)+f(x0)y = f'(x_0)(x - x_0) + f(x_0). So you need two numbers: f(1)f(1) and f′(1)f'(1).

Solutiontry it on your own first

Step 1. Write down the tangent line equation

The tangent line to the graph of ff at the point with xx-coordinate x0x_0 is:

y=f′(x0)(x−x0)+f(x0)y = f'(x_0)(x - x_0) + f(x_0)

With x0=1x_0 = 1:

y=f′(1)(x−1)+f(1)y = f'(1)(x - 1) + f(1)

So we need two numbers: f(1)f(1) and f′(1)f'(1).

Step 2. Work out f(1)

Put x=1x = 1 into the function:

f(1)=13−2⋅1f(1) = 1^3 - 2 \cdot 1
f(1)=1−2f(1) = 1 - 2
f(1)=−1\boxed{f(1) = -1}

So the point of tangency is P=(1,−1)P = (1, -1).

Step 3. Find the derivative f′(x)

Differentiate each term separately (the derivative of a difference is the difference of the derivatives):

(x3)′=3x2,(2x)′=2(x^3)' = 3x^2, \qquad (2x)' = 2

So:

f′(x)=3x2−2\boxed{f'(x) = 3x^2 - 2}

Step 4. Work out f′(1)

Put x=1x = 1 into the derivative:

f′(1)=3⋅12−2f'(1) = 3 \cdot 1^2 - 2
f′(1)=3−2f'(1) = 3 - 2
f′(1)=1\boxed{f'(1) = 1}

This is the gradient of the tangent line: it goes up by 1 for every step of 1 to the right.

Step 5. Substitute into the tangent line equation

We have f(1)=−1f(1) = -1 and f′(1)=1f'(1) = 1:

y=1⋅(x−1)+(−1)y = 1 \cdot (x - 1) + (-1)

Multiply the bracket by 1 and remove it:

y=x−1−1y = x - 1 - 1

Combine the constant terms:

y=x−2y = x - 2

Step 6. Check the result

The tangent line must pass through the point of tangency P=(1,−1)P = (1, -1). Put x=1x = 1:

y=1−2=−1y = 1 - 2 = -1

This matches f(1)=−1f(1) = -1.

Answer

y=x−2\boxed{y = x - 2}

The tangent line to the graph of f(x)=x3−2xf(x) = x^3 - 2x at the point with xx-coordinate x0=1x_0 = 1 is y=x−2y = x - 2.

The theory behind this exerciseCourse: Derivatives from scratch

Sources

We checked the facts on this page against these sources. All of them may be reused freely: they carry an open licence or are not covered by copyright.

  1. Active Calculus (Boelkins) - 1.8 The tangent line approximation (L(x) = f'(a)(x - a) + f(a)) · CC BY-SA 4.0
  2. Active Calculus (Boelkins) - 2.1 Elementary derivative rules (power, constant multiple and sum rules) · CC BY-SA 4.0
  3. Dale Hoffman - Contemporary Calculus, 2.0 Introduction to Derivatives (tangent lines in point-slope form) (PDF) · CC BY 3.0 US
  4. Calaway, Hoffman and Lippman - Applied Calculus, 2.3 Power and Sum Rules for Derivatives (equation of a tangent line) (PDF) · CC BY 3.0 US
  5. Wikibooks - Calculus/Differentiation/Differentiation Defined (slope and equation of the tangent line) · CC BY-SA 4.0
  6. Wikiversity - Mathematics for Applied Sciences (Osnabrück 2023-2024), Part I, Lecture 14: Differentiability (the derivative as the slope of the tangent) · CC BY-SA 4.0

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