Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 50

The graph of the given curve is shown below. Find an expression for its function.




Referring to the graph we can find the equation of the line at left side using point slope form.


$
\begin{equation}
\begin{aligned}
y - 3 &= \frac{0 - 3}{-2+4}(x + 4)\\
\\
y - 3 &= \frac{-3}{2}(x + 4) && (\text{Simplifying equation})\\
\\
2y - 6 &= -3x - 12 && (\text{Solving for }y )\\
\\
y &= \frac{-3}{2}x - 3 \text{ for } -4 \leq x \leq -2\\
\end{aligned}
\end{equation}
$


The next graph is an equation of a semicircle with radius 2.


$
\begin{equation}
\begin{aligned}
y &= \sqrt{4 - x^2} \text{ for } -2 < x < 2\\
\end{aligned}
\end{equation}
$


Again, using point slope form, we can find the equation of the line at the right side.


$
\begin{equation}
\begin{aligned}
y - 0 &= \frac{3-0}{4-2}(x - 2) && (\text{Simplifying equation})\\
\\
y - 0 &= \frac{3}{2} (x -2) && (\text{Solving for }y )\\
\\
y &= \frac{3}{2}x - 3 \text{ for } 2 \leq x \leq 4\\
\end{aligned}
\end{equation}
$



Then, the expression for this function is...
$\boxed{\begin{array}{cccc}
& \frac{-3}{2}x -3& \text{ for } & -4 \leq x \leq -2 \\
&&&\\
f(x) =& \sqrt{4 - x^2} & \text{ for } & -2 < x < 2 \\
&&&\\
& \frac{3}{2}x - 3 & \text{ for } & 2 \leq x \leq 4
\end{array}}$