find the domain and range of the piecewise function graphed below. enter your answers in interval notation…

Explanation:

Step1: Determine the domain

The domain is the set of all x - values for which the function is defined. Looking at the graph:

• The upper piece of the function has a starting point (the filled dot) at (x=-10) and ends at (x = - 8) (the filled dot at (x=-8,y = 2)).
• The lower piece of the function has an open circle at (x=-8) and extends to the right (as the arrow indicates) towards positive infinity. Also, since the upper piece is from (x=-10) to (x=-8) (inclusive of (-10) and (-8)) and the lower piece is from (x>-8) to (\infty), we combine these intervals. The domain is (x\in[-10,\infty)) because the upper part covers from (-10) to (-8) (inclusive) and the lower part covers from (-8) (not inclusive, but since the upper part includes (-8)) to (\infty), so the combined domain is ([-10,\infty)).

Step2: Determine the range

The range is the set of all y - values the function takes.

• For the upper piece (the line from (x = - 10) to (x=-8)): When (x=-10), (y = 8) (from the graph, the upper line starts at ((-10,8))) and ends at (y = 2) (at (x=-8)). So the y - values for the upper piece are from (2) to (8) (inclusive, since the endpoints are filled dots).
• For the lower piece (the line starting at (x=-8) (open circle) and going to the right): The line has a slope that makes the y - values decrease. At (x=-8), the open circle is at (y=-1) (approx, but from the graph, the lower line starts with an open circle at (x = - 8,y=-1)) and goes down towards (-\infty) as (x) increases? Wait, no, looking at the lower line: when (x = 0), (y=-3) (since it crosses the y - axis at ((0,-3))) and as (x) increases, (y) decreases. Wait, actually, let's re - examine the y - values: The upper piece: from (y = 2) (at (x=-8)) up to (y = 8) (at (x=-10)). The lower piece: from (y < 2) (since the open circle at (x=-8) is at a y - value less than (2)) down to (-\infty) (as the arrow on the lower line points downwards as (x) increases? Wait, no, the lower line has an arrow pointing to the right and down, so as (x) approaches (\infty), (y) approaches (-\infty). Also, the upper line's y - values are from (2) to (8) (inclusive), and the lower line's y - values are from (-\infty) up to (2) (not inclusive, but since the upper line includes (2)). Wait, no: Wait, the upper line: when (x=-10), (y = 8); when (x=-8), (y = 2). So the upper line is a decreasing line from ((-10,8)) to ((-8,2)), so the range for the upper line is ([2,8]). The lower line: starts at (x=-8) with an open circle (so (y) at (x=-8) for the lower line is not included, but the value at (x=-8) for the lower line is less than (2), and as (x) increases, the line goes down. So the lower line's y - values go from (-\infty) up to (2) (but not including (2), since the upper line includes (2)). Wait, no, actually, let's check the y - intercept of the lower line. The lower line passes through ((0,-3)) (since when (x = 0), (y=-3)) and has a slope. Let's calculate the slope: from (x=-8) (open circle) to (x = 0), the change in (x) is (8), change in (y) is (-3-\left(-1\right)=-2) (assuming the open circle at (x=-8) is (y=-1)). So slope (m=\frac{-3 - (-1)}{0 - (-8)}=\frac{-2}{8}=-\frac{1}{4}). So the equation of the lower line is (y - (-1)=-\frac{1}{4}(x + 8)), (y+1=-\frac{1}{4}x - 2), (y=-\frac{1}{4}x-3). As (x) approaches (\infty), (y) approaches (-\infty), and as (x) approaches (-8) from the right, (y) approaches (-\frac{1}{4}(-8)-3=2 - 3=-1). So the lower line's y - values are ((-\infty,2)) (since at (x=-8), the open circle is at (y=-1) which is less than (2), and as (x) increases, (y) decreases to (-\infty)). The upper line's y - values are ([2,8]). So combining these, the range is ((-\infty,8]) because the lower line covers from (-\infty) to (2) (not inclusive) and the upper line covers from (2) (inclusive) to (8) (inclusive), so the combined range is ((-\infty,8]).

Answer:

Domain: (\boldsymbol{[-10,\infty)}) Range: (\boldsymbol{(-\infty,8]})