Vtyjkyuk

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=$frac0-2-1+2$=-2

d=$sqrt((0-2)^2+(-1+2)^2)$=$sqrt5$

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?

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 $y=sqrt1-x^2$. Accordingly the function can be expressed in the form

beginequation
f(x)=begincases
-2x-2text for -2le x<-1\
sqrt1-x^2text for -1le xle1
endcases
endequation

Another way to express the function is using the $textbfunit step function$

beginequation
U(x)=begincases
0text for x<0\
1text for xge0
endcases
endequation

Then

begineqnarray
f(x)&=&-2(x+1)U(x+2)+left[sqrt1-x^2+2(x+1)right],U(x+1)\
&-&sqrt1-x^2U(x-1)
endeqnarray

However, this assumes it is defined everywhere on $mathbbR$ and equal to $0$ outside the given domain.

Here is a graph: