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

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?

Answer 1

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-2\text{ for }-2\le x<-1\\ \sqrt{1-x^2}\text{ for }-1\le x\le1 \end{cases} \end{equation}

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

\begin{equation} U(x)=\begin{cases} 0\text{ for }x<0\\ 1\text{ for }x\ge0 \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: