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Rational functions: parameters and tangents

Rational functions: parameters and tangents

A family with a parameter

A letter in the formula besides x\displaystyle x – a parameter – makes a family of functions, one for each value of the letter. A fact about one member of the family is an equation in the parameter. The example below works with the family f(x)=x2+ax\displaystyle {f(x)}={\frac{x^2+a}{x}}:

  • f′(x)=2x⋅x−(x2+a)x2=x2−ax2\displaystyle {f'(x)}={\frac{2x\cdot x-(x^2+a)}{x^2}}={\frac{x^2-a}{x^2}}.
  • For an extremum point at x=2\displaystyle {x}={2}: f′(2)=0\displaystyle {f'(2)}={0}, so 4−a4=0\displaystyle {\frac{4-a}{4}}={0} and a=4\displaystyle {a}={4}.
  • For a=4\displaystyle {a}={4}: f(x)=x2+4x\displaystyle {f(x)}={\frac{x^2+4}{x}}, with a minimum point (2,4)\displaystyle (2,4) and a maximum point (−2,−4)\displaystyle (-2,-4).
  • For a<0\displaystyle a<0: x2−a>0\displaystyle x^2-a>0 for every x\displaystyle x, so f′(x)>0\displaystyle f'(x)>0 throughout the domain – the function rises, and has no extremum points.
The tangent at a point of the graph

The tangent at the point of the graph where x=x1\displaystyle {x}={x_1} is the line through (x1,f(x1))\displaystyle (x_1,f(x_1)) with the slope f′(x1)\displaystyle f'(x_1). The example below finds the tangent to f(x)=4x\displaystyle {f(x)}={\frac{4}{x}} at the point where x=2\displaystyle {x}={2}:

  • f(2)=2\displaystyle {f(2)}={2}: the point (2,2)\displaystyle (2,2);
  • f′(x)=−4x2\displaystyle {f'(x)}={-\frac{4}{x^2}}, so f′(2)=−1\displaystyle {f'(2)}={-1};
  • y−2=−1⋅(x−2)\displaystyle {y-2}={-1\cdot(x-2)}, that is y=−x+4\displaystyle {y}={-x+4}.
f(x)=4x\displaystyle f(x)=\frac{4}{x}
y=−x+4\displaystyle y=-x+4
(2,2)\displaystyle (2,2)
Tangents with a given slope

When the slope of the tangent is given and the point is not, the equation f′(x)=m\displaystyle {f'(x)}={m} gives the points. The example below finds the tangents to f(x)=xx−1\displaystyle {f(x)}={\frac{x}{x-1}} whose slope is −1\displaystyle -1:

  • f′(x)=(x−1)−x(x−1)2=−1(x−1)2\displaystyle {f'(x)}={\frac{(x-1)-x}{(x-1)^2}}={\frac{-1}{(x-1)^2}};
  • −1(x−1)2=−1\displaystyle {\frac{-1}{(x-1)^2}}={-1} gives (x−1)2=1\displaystyle {(x-1)^2}={1}, so x=0\displaystyle {x}={0} or x=2\displaystyle {x}={2};
  • the points (0,0)\displaystyle (0,0) and (2,2)\displaystyle (2,2), and the tangents y=−x\displaystyle {y}={-x} and y=−x+4\displaystyle {y}={-x+4}.
f(x)=xx−1\displaystyle f(x)=\frac{x}{x-1}
x=1\displaystyle x=1
y=1\displaystyle y=1
y=−x\displaystyle y=-x
y=−x+4\displaystyle y=-x+4

The two tangents are parallel: the graph of f\displaystyle f is symmetric about the point (1,1)\displaystyle (1,1) where its asymptotes meet.

The asymptotes of a moved graph

The asymptotes move with the graph:

  • g(x)=f(x−p)+k\displaystyle {g(x)}={f(x-p)+k}: the graph moves p\displaystyle p to the right and k\displaystyle k up, so an asymptote x=a\displaystyle {x}={a} becomes x=a+p\displaystyle {x}={a+p}, and y=b\displaystyle {y}={b} becomes y=b+k\displaystyle {y}={b+k};
  • g(x)=∣f(x)∣\displaystyle {g(x)}={|f(x)|}: the parts of the graph under the x\displaystyle x-axis flip above it; a vertical asymptote stays, and y=b\displaystyle {y}={b} becomes y=∣b∣\displaystyle {y}={|b|}.
Exercise 1

The family is f(x)=axx2+4\displaystyle {f(x)}={\frac{ax}{x^2+4}}, with a≠0\displaystyle a\neq0.
One of the functions has a maximum point at (2,1)\displaystyle (2,1).

a

Find f′(x)\displaystyle f'(x).

f(x)=axx2+4\displaystyle f(x)=\frac{ax}{x^{2}+4}

    / – 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

    Explain why every function in the family has extremum points at x=±2\displaystyle {x}={\pm2}.

    c

    f(2)=2a8\displaystyle {f(2)}={\frac{2a}{8}}.
    Find a\displaystyle a.

      d

      Mark the extremum points of the function you found.

      f(x)=4xx2+4\displaystyle f(x)=\frac{4x}{x^{2}+4}

      Exercise 2

      The function is f(x)=x+1x−1\displaystyle {f(x)}={\frac{x+1}{x-1}}.

      a

      Find f′(x)\displaystyle f'(x).

      f(x)=x+1x−1\displaystyle f(x)=\frac{x+1}{x-1}

        / – 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 on the board the equation of the tangent at the point where x=3\displaystyle {x}={3}.

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

        c

        At which points of the graph is the slope of the tangent −2\displaystyle -2?

          d

          Write on the board the equations of the two tangents whose slope is −2\displaystyle -2.

          f(x)=x+1x−1\displaystyle f(x)=\frac{x+1}{x-1}
          x=1\displaystyle x=1

          Exercise 3

          The family is f(x)=x2+ax−1\displaystyle {f(x)}={\frac{x^2+a}{x-1}}.

          a

          Find f′(x)\displaystyle f'(x).

          f(x)=x2+ax−1\displaystyle f(x)=\frac{x^{2}+a}{x-1}

            / – 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

            One function of the family has an extremum point at x=3\displaystyle {x}={3}.
            f′(3)=9−6−a4\displaystyle {f'(3)}={\frac{9-6-a}{4}}.
            Find a\displaystyle a.

              c

              f\displaystyle f has no extremum points when the numerator of f′\displaystyle f', x2−2x−a\displaystyle x^2-2x-a, has no roots apart from a root it shares with the denominator.
              Its discriminant is 4+4a\displaystyle 4+4a.
              Solve the inequality.

                d

                What is the graph of f\displaystyle f for a=−1\displaystyle {a}={-1}? Explain why it has no extremum points.

                Exercise 4

                The function is f(x)=2x\displaystyle {f(x)}={\frac{2}{x}}.
                g(x)=f(x−1)+3\displaystyle {g(x)}={f(x-1)+3}.

                a

                Write on the board the equations of the asymptotes of g\displaystyle g.

                g(x)=2x−1+3\displaystyle g(x)=\frac{2}{x-1}+3

                b

                What are the asymptotes of h(x)=∣f(x)−3∣\displaystyle {h(x)}={|f(x)-3|}? Explain.