Answer:
[tex]m(x)=\begin{cases}-\dfrac{1}{3}(x+4)^2+3&\text{ for }x \le-1\\4\left(\dfrac{1}{2}\right)^x&\text{ for }-1 < x < 3\\-x+5&\text{ for }3\le x\end{cases}[/tex]
Step-by-step explanation:
The domain is the set of x-values for which the function is applicable. The pieces of a piecewise-defined function are each defined on their own domain. Here, there are three different functions, each defined on a different domain.
From left to right, the function's domain can be divided into the sections ...
x ≤ -1
This section of the graph looks like a parabola that opens downward. Its vertex is (-4, 3), and it seems to have a scale factor less than 1.
For some scale factor 'a', the function in vertex form is ...
y = a(x -h)² +k . . . . . . quadratic with vertex (h, k)
y = a(x +4)² +3
We know the point (x, y) = (-1, 0) is on the curve, so we can use these values to find 'a':
0 = a(-1 +4)² +3 = 9a +3
a = -3/9 = -1/3
So, the left-section function is ...
m(x) = -1/3(x +4)² +3
__
-1 < x < 3
The middle section of the graph has increasing slope, so might be a parabola or an exponential function. We note the average rate of change goes from -4 in the interval (-1, 0) to -2 in the interval (0, 1) to -1 in the interval (1, 2). The slope changing by a constant factor (1/2) in each unit interval is characteristic of an exponential function. That factor is the base of the exponent.
The actual values on the curve also decrease by a factor of 1/2 in each unit interval, which tells us the function has not been translated vertically. The y-intercept value of 4 at x=0 tells us the multiplier of the function:
m(x) = 4(1/2)^x
__
3 ≤ x
The funciton in this domain is a straight line. It has a "rise" of -1 unit for each "run" of 1 unit, so its slope is -1. If we extend the line left to the y-axis, we see that it has a y-intercept of 5. Its equation is ...
m(x) = -x +5
__
Putting the pieces together into one function description, we have ...
[tex]m(x)=\begin{cases}-\dfrac{1}{3}(x+4)^2+3&\text{ for }x \le-1\\4\left(\dfrac{1}{2}\right)^x&\text{ for }-1 < x < 3\\-x+5&\text{ for }3\le x\end{cases}[/tex]
_____
Additional comment
If the function in the middle section were quadratic, its average rate of change on adjacent equal intervals would form an arithmetic sequence. Because the sequence is geometric, we know it is an exponential function.
