Skip to content

Practice: one over a function and transformations

Practice: one over a function and transformations

1

f(x)=x2−6x+8\displaystyle {f(x)}={x^2-6x+8}, and g(x)=1f(x)\displaystyle {g(x)}={\frac{1}{f(x)}}.

a

Solve the equation f(x)=0\displaystyle {f(x)}={0}, and write the domain of g\displaystyle g.

    b

    Write the equations of the asymptotes of g\displaystyle g on the board, and mark its extremum point.

    f(x)=x2−6x+8\displaystyle f(x)=x^{2}-6x+8
    g(x)=1x2−6x+8\displaystyle g(x)=\frac{1}{x^{2}-6x+8}

    c

    On which intervals is g\displaystyle g negative? On which intervals does it rise?

    2

    f(x)=x2+4x+5\displaystyle {f(x)}={x^2+4x+5}, and g(x)=1f(x)\displaystyle {g(x)}={\frac{1}{f(x)}}.

    a

    Mark the extremum point of g\displaystyle g.

    g(x)=1x2+4x+5\displaystyle g(x)=\frac{1}{x^{2}+4x+5}

    b

    For which values of x\displaystyle x is g(x)=12\displaystyle {g(x)}={\frac{1}{2}}?

      c

      What is the domain of g\displaystyle g? How many asymptotes does it have, and what are they? On which intervals does it rise, and on which does it fall?

      3

      On the board – the graph of f(x)=1x2\displaystyle {f(x)}={\frac{1}{x^2}}.
      It is moved 1 unit to the left and 3 units down.
      Write the new function on the board, and the equations of its asymptotes.

      f(x)=1x2\displaystyle f(x)=\frac{1}{x^{2}}

      4

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

      a

      By which operations is g\displaystyle g made from the function 1x\displaystyle \frac{1}{x}?

      b

      Write the equations of the asymptotes of g\displaystyle g on the board, and mark the point where the graph crosses the x\displaystyle x-axis.

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

      5

      On the board – the graph of g(x)=1x2−4\displaystyle {g(x)}={\frac{1}{x^2-4}}, and beside it the graph of h(x)\displaystyle h(x), made from g\displaystyle g by one transformation.
      Write h(x)\displaystyle h(x) on the board.

      g(x)=1x2−4\displaystyle g(x)=\frac{1}{x^{2}-4}
      h(x)= ?\displaystyle h(x)=\,?

      6

      g(x)=1x2−1\displaystyle {g(x)}={\frac{1}{x^2-1}}.
      Solve the equation ∣g(x)∣=1\displaystyle {\left|g(x)\right|}={1}.