An expression for the function whose graph consists of the line segment from point to point












0















An expression for the function whose graph consists of the line segment from (-2,2) and (-1,0) together with the top half of the circle with center the origin and radius 1.



m=$frac{0-2}{-1+2}$=-2

d=${sqrt{((0-2)^2+(-1+2)^2)}}$=${sqrt{5}}$

y=2x-2

${x^2+y^2=r^2}$

${x^2+y^2=1}$

${y^2=1-x^2}$

${y=sqrt{(1-x^2)}}$



do i need to add something more?







algebra-precalculus







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  • It should be $y=-2x-2$. See my answer below.
    – John Wayland Bales
    Nov 13 '16 at 19:24
















0















An expression for the function whose graph consists of the line segment from (-2,2) and (-1,0) together with the top half of the circle with center the origin and radius 1.



m=$frac{0-2}{-1+2}$=-2

d=${sqrt{((0-2)^2+(-1+2)^2)}}$=${sqrt{5}}$

y=2x-2

${x^2+y^2=r^2}$

${x^2+y^2=1}$

${y^2=1-x^2}$

${y=sqrt{(1-x^2)}}$



do i need to add something more?







algebra-precalculus







share|cite|improve this question






















  • It should be $y=-2x-2$. See my answer below.
    – John Wayland Bales
    Nov 13 '16 at 19:24














0












0








0








An expression for the function whose graph consists of the line segment from (-2,2) and (-1,0) together with the top half of the circle with center the origin and radius 1.



m=$frac{0-2}{-1+2}$=-2

d=${sqrt{((0-2)^2+(-1+2)^2)}}$=${sqrt{5}}$

y=2x-2

${x^2+y^2=r^2}$

${x^2+y^2=1}$

${y^2=1-x^2}$

${y=sqrt{(1-x^2)}}$



do i need to add something more?







algebra-precalculus







share|cite|improve this question














An expression for the function whose graph consists of the line segment from (-2,2) and (-1,0) together with the top half of the circle with center the origin and radius 1.



m=$frac{0-2}{-1+2}$=-2

d=${sqrt{((0-2)^2+(-1+2)^2)}}$=${sqrt{5}}$

y=2x-2

${x^2+y^2=r^2}$

${x^2+y^2=1}$

${y^2=1-x^2}$

${y=sqrt{(1-x^2)}}$



do i need to add something more?







algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











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asked Nov 13 '16 at 16:14









Alyssa xAlyssa x

6817




6817












  • It should be $y=-2x-2$. See my answer below.
    – John Wayland Bales
    Nov 13 '16 at 19:24


















  • It should be $y=-2x-2$. See my answer below.
    – John Wayland Bales
    Nov 13 '16 at 19:24
















It should be $y=-2x-2$. See my answer below.
– John Wayland Bales
Nov 13 '16 at 19:24




It should be $y=-2x-2$. See my answer below.
– John Wayland Bales
Nov 13 '16 at 19:24










1 Answer
1






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oldest

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0














This is an example of a "piecewise defined function."



On the interval $[-2,-1]$ the function is defined by $y=-2x-2$ whereas on the interval $[-1,1]$ it is defined by the $y=sqrt{1-x^2}$. Accordingly the function can be expressed in the form



begin{equation}
f(x)=begin{cases}
-2x-2text{ for }-2le x<-1\
sqrt{1-x^2}text{ for }-1le xle1
end{cases}
end{equation}



Another way to express the function is using the $textbf{unit step function}$



begin{equation}
U(x)=begin{cases}
0text{ for }x<0\
1text{ for }xge0
end{cases}
end{equation}



Then



begin{eqnarray}
f(x)&=&-2(x+1)U(x+2)+left[sqrt{1-x^2}+2(x+1)right],U(x+1)\
&-&sqrt{1-x^2}U(x-1)
end{eqnarray}



However, this assumes it is defined everywhere on $mathbb{R}$ and equal to $0$ outside the given domain.



Here is a graph:



Graph of piecewise function






share|cite|improve this answer























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    1 Answer
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    1 Answer
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    active

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    0














    This is an example of a "piecewise defined function."



    On the interval $[-2,-1]$ the function is defined by $y=-2x-2$ whereas on the interval $[-1,1]$ it is defined by the $y=sqrt{1-x^2}$. Accordingly the function can be expressed in the form



    begin{equation}
    f(x)=begin{cases}
    -2x-2text{ for }-2le x<-1\
    sqrt{1-x^2}text{ for }-1le xle1
    end{cases}
    end{equation}



    Another way to express the function is using the $textbf{unit step function}$



    begin{equation}
    U(x)=begin{cases}
    0text{ for }x<0\
    1text{ for }xge0
    end{cases}
    end{equation}



    Then



    begin{eqnarray}
    f(x)&=&-2(x+1)U(x+2)+left[sqrt{1-x^2}+2(x+1)right],U(x+1)\
    &-&sqrt{1-x^2}U(x-1)
    end{eqnarray}



    However, this assumes it is defined everywhere on $mathbb{R}$ and equal to $0$ outside the given domain.



    Here is a graph:



    Graph of piecewise function






    share|cite|improve this answer




























      0














      This is an example of a "piecewise defined function."



      On the interval $[-2,-1]$ the function is defined by $y=-2x-2$ whereas on the interval $[-1,1]$ it is defined by the $y=sqrt{1-x^2}$. Accordingly the function can be expressed in the form



      begin{equation}
      f(x)=begin{cases}
      -2x-2text{ for }-2le x<-1\
      sqrt{1-x^2}text{ for }-1le xle1
      end{cases}
      end{equation}



      Another way to express the function is using the $textbf{unit step function}$



      begin{equation}
      U(x)=begin{cases}
      0text{ for }x<0\
      1text{ for }xge0
      end{cases}
      end{equation}



      Then



      begin{eqnarray}
      f(x)&=&-2(x+1)U(x+2)+left[sqrt{1-x^2}+2(x+1)right],U(x+1)\
      &-&sqrt{1-x^2}U(x-1)
      end{eqnarray}



      However, this assumes it is defined everywhere on $mathbb{R}$ and equal to $0$ outside the given domain.



      Here is a graph:



      Graph of piecewise function






      share|cite|improve this answer


























        0












        0








        0






        This is an example of a "piecewise defined function."



        On the interval $[-2,-1]$ the function is defined by $y=-2x-2$ whereas on the interval $[-1,1]$ it is defined by the $y=sqrt{1-x^2}$. Accordingly the function can be expressed in the form



        begin{equation}
        f(x)=begin{cases}
        -2x-2text{ for }-2le x<-1\
        sqrt{1-x^2}text{ for }-1le xle1
        end{cases}
        end{equation}



        Another way to express the function is using the $textbf{unit step function}$



        begin{equation}
        U(x)=begin{cases}
        0text{ for }x<0\
        1text{ for }xge0
        end{cases}
        end{equation}



        Then



        begin{eqnarray}
        f(x)&=&-2(x+1)U(x+2)+left[sqrt{1-x^2}+2(x+1)right],U(x+1)\
        &-&sqrt{1-x^2}U(x-1)
        end{eqnarray}



        However, this assumes it is defined everywhere on $mathbb{R}$ and equal to $0$ outside the given domain.



        Here is a graph:



        Graph of piecewise function






        share|cite|improve this answer














        This is an example of a "piecewise defined function."



        On the interval $[-2,-1]$ the function is defined by $y=-2x-2$ whereas on the interval $[-1,1]$ it is defined by the $y=sqrt{1-x^2}$. Accordingly the function can be expressed in the form



        begin{equation}
        f(x)=begin{cases}
        -2x-2text{ for }-2le x<-1\
        sqrt{1-x^2}text{ for }-1le xle1
        end{cases}
        end{equation}



        Another way to express the function is using the $textbf{unit step function}$



        begin{equation}
        U(x)=begin{cases}
        0text{ for }x<0\
        1text{ for }xge0
        end{cases}
        end{equation}



        Then



        begin{eqnarray}
        f(x)&=&-2(x+1)U(x+2)+left[sqrt{1-x^2}+2(x+1)right],U(x+1)\
        &-&sqrt{1-x^2}U(x-1)
        end{eqnarray}



        However, this assumes it is defined everywhere on $mathbb{R}$ and equal to $0$ outside the given domain.



        Here is a graph:



        Graph of piecewise function







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 13 '16 at 19:43

























        answered Nov 13 '16 at 19:22









        John Wayland BalesJohn Wayland Bales

        13.9k21238




        13.9k21238






























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