Derivatives: the basics

Equation of a tangent line

The tangent line to y=f(x)y=f(x) at x=ax=a is the line with slope f′(a)f'(a) through the point (a, f(a))(a,\,f(a)).

y=f′(a)(x−a)+f(a)y=f'(a)(x-a)+f(a)

A tangent line is the line with slope f′(a)f'(a) through the point (a, f(a))(a,\,f(a)).

y=f′(a)(x−a)+f(a)y=f'(a)(x-a)+f(a)

Get the slope by differentiating, plug in the coordinates of the point, and the equation is determined. Do not forget to compute the yy-coordinate of the point of tangency from f(a)f(a).

xyy = f(x)a(a, f(a))
The tangent line passes through (a, f(a)) with slope f'(a).Drag the dot

Examples

  1. 01Tangent line at x=2x=2. What goes in □?f(x)=x2  ⇒  y=□(x−2)+4\displaystyle f(x)=x^{2} \;\Rightarrow\; y=\square(x-2)+4AnswerHide
    Answer4\displaystyle 4

    Steps

    1. □ is the tangent slope: the value of the derivative at the point
    2. f′(x)\displaystyle f'(x)=2x\displaystyle = 2xf′(2)\displaystyle f'(2)=4\displaystyle = 4
  2. 02What are the coordinates of the point of tangency at x=1x=1?f(x)=x2\displaystyle f(x)=x^{2}AnswerHide
    Answer(1, 1)\displaystyle (1,\ 1)

    Steps

    1. The point sits on the curve: its xx is 11 and its height comes from ff
    2. f(1)\displaystyle f(1)=12=1\displaystyle = 1^{2} = 1
    3. So the point of tangency is (1, 1)(1,\ 1)
  3. 03Tangent line at x=−1x=-1. What goes in □?f(x)=x2  ⇒  y=□(x+1)+1\displaystyle f(x)=x^{2} \;\Rightarrow\; y=\square(x+1)+1AnswerHide
    Answer−2\displaystyle -2

    Steps

    1. □ is the tangent slope: the value of the derivative at the point
    2. f′(x)\displaystyle f'(x)=2x\displaystyle = 2xf′(−1)\displaystyle f'(-1)=−2\displaystyle = -2
  4. 04Tangent line at x=1x=1. What goes in □?f(x)=x2  ⇒  y=□(x−1)+1\displaystyle f(x)=x^{2} \;\Rightarrow\; y=\square(x-1)+1AnswerHide
    Answer2\displaystyle 2

    Steps

    1. □ is the tangent slope: the value of the derivative at the point
    2. f′(x)\displaystyle f'(x)=2x\displaystyle = 2xf′(1)\displaystyle f'(1)=2\displaystyle = 2
  5. 05Tangent line at x=3x=3. What goes in □?f(x)=x2  ⇒  y=□(x−3)+9\displaystyle f(x)=x^{2} \;\Rightarrow\; y=\square(x-3)+9AnswerHide
    Answer6\displaystyle 6

    Steps

    1. □ is the tangent slope: the value of the derivative at the point
    2. f′(x)\displaystyle f'(x)=2x\displaystyle = 2xf′(3)\displaystyle f'(3)=6\displaystyle = 6
  6. 06Tangent line at x=−2x=-2. What goes in □?f(x)=x2  ⇒  y=□(x+2)+4\displaystyle f(x)=x^{2} \;\Rightarrow\; y=\square(x+2)+4AnswerHide
    Answer−4\displaystyle -4

    Steps

    1. □ is the tangent slope: the value of the derivative at the point
    2. f′(x)\displaystyle f'(x)=2x\displaystyle = 2xf′(−2)\displaystyle f'(-2)=−4\displaystyle = -4

FAQ

Where does the slope of the tangent come from?
From the derivative at the point of tangency, f′(a)f'(a). The steps are: find f′(x)f'(x), substitute x=ax=a, then write the line from the point and the slope.

Published 2026-09-03 · Updated 2026-09-05