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The equation of the tangent

The equation of the tangent

The tangent at a point on the graph

The tangent to the graph of f\displaystyle f at the point where x=x1\displaystyle {x}={x_1} is a line through the point (x1,y1)\displaystyle (x_1,y_1), and its slope is the derivative there:

m\displaystyle m =\displaystyle = f′(x1)\displaystyle f'(x_1)

A line of slope m\displaystyle m through (x1,y1)\displaystyle (x_1,y_1):

y−y1\displaystyle y-y_1 =\displaystyle = m(x−x1)\displaystyle m(x-x_1)

So the equation of the tangent is found in three steps:

  • y1=f(x1)\displaystyle {y_1}={f(x_1)} – the y\displaystyle y-coordinate of the point;
  • m=f′(x1)\displaystyle {m}={f'(x_1)} – the slope;
  • substitute in y−y1=m(x−x1)\displaystyle {y-y_1}={m(x-x_1)} and bring it to the form y=mx+b\displaystyle {y}={mx+b}.
Example: f(x)=x3−2x+1\displaystyle {f(x)}={x^3-2x+1} at x=1\displaystyle {x}={1}

The example below finds the equation of the tangent to the graph of f(x)=x3−2x+1\displaystyle {f(x)}={x^3-2x+1} at the point where x=1\displaystyle {x}={1}.

  • y1=f(1)=1−2+1=0\displaystyle {y_1}={f(1)}={1-2+1}={0}: the point is (1,0)\displaystyle (1,0).
  • f′(x)=3x2−2\displaystyle {f'(x)}={3x^2-2}, so m=f′(1)=3−2=1\displaystyle {m}={f'(1)}={3-2}={1}.
  • y−0=1⋅(x−1)\displaystyle {y-0}={1\cdot(x-1)}, that is y=x−1\displaystyle {y}={x-1}.
f(x)=x3−2x+1\displaystyle f(x)=x^{3}-2x+1
y=x−1\displaystyle y=x-1
A=(1,0)\displaystyle A=(1,0)

Near (1,0)\displaystyle (1,0) the tangent almost merges with the graph. Far from the point it may cut the graph again – here at x=−2\displaystyle {x}={-2}, at the point (−2,−3)\displaystyle (-2,-3).

A tangent of a given slope

When the slope of the tangent is given and not the point, the slope fixes the point: solve the equation f′(x)=m\displaystyle {f'(x)}={m}.
Parallel lines are lines with the same slope, so a tangent parallel to a given line is a tangent with the slope of that line.

The example below finds the tangent to the graph of f(x)=x2−6x+5\displaystyle {f(x)}={x^2-6x+5} parallel to the line y=2x+3\displaystyle {y}={2x+3}.

  • The slope of the line is 2, so solve f′(x)=2\displaystyle {f'(x)}={2}: 2x−6=2\displaystyle {2x-6}={2}, that is x=4\displaystyle {x}={4}.
  • y1=f(4)=16−24+5=−3\displaystyle {y_1}={f(4)}={16-24+5}={-3}: the point is (4,−3)\displaystyle (4,-3).
  • y+3=2(x−4)\displaystyle {y+3}={2(x-4)}, that is y=2x−11\displaystyle {y}={2x-11}.
f(x)=x2−6x+5\displaystyle f(x)=x^{2}-6x+5
y=2x+3\displaystyle y=2x+3
y=2x−11\displaystyle y=2x-11
A=(4,−3)\displaystyle A=(4,-3)
A horizontal tangent, and a tangent and the axes
  • A horizontal tangent: its slope is 0, so it is at a point where f′(x)=0\displaystyle {f'(x)}={0}. Its equation is y=y1\displaystyle {y}={y_1}. On a parabola – the tangent at the vertex.
  • Where the tangent meets the axes: the tangent is a straight line, so substitute x=0\displaystyle {x}={0} in it (where it meets the y\displaystyle y-axis) or y=0\displaystyle {y}={0} (where it meets the x\displaystyle x-axis).

For example, the tangent y=2x−11\displaystyle {y}={2x-11} meets the y\displaystyle y-axis at (0,−11)\displaystyle (0,-11), and the x\displaystyle x-axis where 2x−11=0\displaystyle {2x-11}={0}, that is at (5.5,0)\displaystyle (5.5,0).

Exercise 1

The function is f(x)=x2−3x+2\displaystyle {f(x)}={x^2-3x+2}.
The point (3,2)\displaystyle (3,2) is on its graph.

a

Work out the slope of the tangent to the graph at the point (3,2)\displaystyle (3,2).

f(x)=x2−3x+2\displaystyle f(x)=x^{2}-3x+2

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b

    Write the equation of the tangent on the board.

    f(x)=x2−3x+2\displaystyle f(x)=x^{2}-3x+2
    (3,2)\displaystyle (3,2)

    Exercise 2

    The function is f(x)=−x3+3x2\displaystyle {f(x)}={-x^3+3x^2}.
    Write on the board the equation of the tangent to the graph of f\displaystyle f at the point where x=1\displaystyle {x}={1}.

    f(x)=−x3+3x2\displaystyle f(x)=-x^{3}+3x^{2}
    (1,2)\displaystyle (1,2)

    Exercise 3

    The function is f(x)=x3−3x2\displaystyle {f(x)}={x^3-3x^2}.
    Its graph has two tangents of slope 9.

    a

    Write an equation and solve it: at which values of x\displaystyle x is the slope of the tangent 9?

      b

      Write the equations of both tangents on the board.

      f(x)=x3−3x2\displaystyle f(x)=x^{3}-3x^{2}

      Exercise 4

      The function is f(x)=x2+4x\displaystyle {f(x)}={x^2+4x}.

      a

      At which value of x\displaystyle x is the tangent to the graph horizontal?
      Write an equation and solve it.

        b

        Write the equation of the horizontal tangent on the board.

        f(x)=x2+4x\displaystyle f(x)=x^{2}+4x

        Exercise 5

        The function is f(x)=x2−4\displaystyle {f(x)}={x^2-4}.

        a

        Write on the board the equation of the tangent to the graph at the point (1,−3)\displaystyle (1,-3).

        f(x)=x2−4\displaystyle f(x)=x^{2}-4
        (1,−3)\displaystyle (1,-3)

        b

        Find the x\displaystyle x-coordinate of the point where the tangent meets the x\displaystyle x-axis.

          c

          The tangent meets the y\displaystyle y-axis at (0,−5)\displaystyle (0,-5).
          Work out the area of the triangle the tangent makes with the axes.

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

            Exercise 6

            The function is f(x)=x2+bx\displaystyle {f(x)}={x^2+bx} (b\displaystyle b a parameter).
            The tangent to the graph at the point where x=1\displaystyle {x}={1} is parallel to the line y=5x\displaystyle {y}={5x}.

            a

            f′(x)=2x+b\displaystyle {f'(x)}={2x+b}.
            Write an equation in b\displaystyle b and solve it.

              b

              Write the equation of the tangent.