An expression for the function whose graph consists of the line segment from point to point
$begingroup$
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
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
$endgroup$
add a comment |
$begingroup$
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
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
$endgroup$
$begingroup$
It should be $y=-2x-2$. See my answer below.
$endgroup$
– John Wayland Bales
Nov 13 '16 at 19:24
add a comment |
$begingroup$
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
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
$endgroup$
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
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
asked Nov 13 '16 at 16:14
Alyssa xAlyssa x
6817
6817
$begingroup$
It should be $y=-2x-2$. See my answer below.
$endgroup$
– John Wayland Bales
Nov 13 '16 at 19:24
add a comment |
$begingroup$
It should be $y=-2x-2$. See my answer below.
$endgroup$
– John Wayland Bales
Nov 13 '16 at 19:24
$begingroup$
It should be $y=-2x-2$. See my answer below.
$endgroup$
– John Wayland Bales
Nov 13 '16 at 19:24
$begingroup$
It should be $y=-2x-2$. See my answer below.
$endgroup$
– John Wayland Bales
Nov 13 '16 at 19:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
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:
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
add a comment |
$begingroup$
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:
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
add a comment |
$begingroup$
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:
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
$endgroup$
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:
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
edited Nov 13 '16 at 19:43
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
answered Nov 13 '16 at 19:22
John Wayland BalesJohn Wayland Bales
13.9k21238
13.9k21238
add a comment |
add a comment |
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$begingroup$
It should be $y=-2x-2$. See my answer below.
$endgroup$
– John Wayland Bales
Nov 13 '16 at 19:24