Question

evaluating a piecewise function using the graph a linear piecewise function is represented by the graph. use the graph to evaluate the function. when x = -5, y = dropdown. when x = -1, y = dropdown. when x = 3, y = dropdown.

Explanation:

Step1: Evaluate at \( x = -5 \)

The left - most horizontal line (red) has a \( y \) - value of 3 (since the open circle is at \( (-1,3) \) and it extends to the left). For \( x=-5 \), we look at this horizontal line. So when \( x = - 5 \), \( y=3 \).

Step2: Evaluate at \( x=-1 \)

We need to check the pieces of the piece - wise function. The horizontal line has an open circle at \( x = - 1 \), so we look at the other piece (the line with the closed circle at \( (-1, - 2) \)? Wait, no, looking at the graph, the lower line has a closed circle at \( x=-1,y = - 2 \)? Wait, no, let's re - examine. Wait, the upper horizontal line (red) has an open circle at \( x=-1,y = 3 \), and the lower line (the non - horizontal line) has a closed circle at \( x=-1,y=-2 \)? Wait, no, the grid: let's count the grid. The upper horizontal line: each grid square is 1 unit. The upper line is at \( y = 3 \) (since from the origin, up 3 units). The open circle is at \( x=-1,y = 3 \). The lower line: the closed circle is at \( x=-1,y=-2 \)? Wait, no, the closed circle is at \( x=-1,y=-2 \)? Wait, the lower line (the line with the closed circle) passes through \( (-1,-2) \) and then goes to the right. Wait, when \( x=-1 \), we have to see which piece includes \( x=-1 \). The open circle means the upper piece does not include \( x=-1 \), so we take the lower piece. Wait, no, maybe I misread. Wait, the upper horizontal line: direction is left, open circle at \( (-1,3) \). The lower line: the closed circle is at \( (-1, - 2) \)? Wait, no, the y - axis: the upper horizontal line is at \( y = 3 \) (since between \( y = 2 \) and \( y = 4 \), so \( y=3 \)). The lower line: the closed circle is at \( x=-1,y=-2 \)? Wait, no, the grid: let's see, the lower line (the line with the closed circle) has a point at \( x=-1,y=-2 \) and then goes to the right. Wait, when \( x=-1 \), since the upper line has an open circle at \( x=-1 \), we use the lower line. So when \( x=-1 \), \( y=-2 \)? Wait, no, maybe I made a mistake. Wait, the problem says "linear piecewise function". Let's re - check:

Wait, the upper horizontal line: for \( x < - 1 \), the function is \( y = 3 \) (since the open circle is at \( x=-1,y = 3 \), so the domain of the upper line is \( x < - 1 \)). For \( x\geq - 1 \), we use the other line. The other line (the non - horizontal line) has a closed circle at \( x=-1 \). Let's find the equation of the lower line. The closed circle is at \( (-1,-2) \), and let's find another point. When \( x = 3 \), let's see, if we go from \( x=-1,y=-2 \) to \( x = 3 \), how much does \( y \) change? Let's assume the slope. Wait, maybe it's easier:

For \( x=-5 \): \( x=-5 < - 1 \), so we use the upper horizontal line. So \( y = 3 \).

For \( x=-1 \): \( x=-1 \) is the boundary. The upper line has an open circle at \( x=-1 \), so we use the lower line. The lower line has a closed circle at \( x=-1 \), so we look at the lower line. Let's find the value. The lower line: when \( x=-1 \), \( y=-2 \)? Wait, no, the closed circle is at \( x=-1,y=-2 \)? Wait, the grid: the closed circle is at \( (-1,-2) \), yes. So when \( x=-1 \), \( y=-2 \).

For \( x = 3 \): \( x = 3\geq - 1 \), so we use the lower line. Let's find the equation of the lower line. The two points on the lower line: \( (-1,-2) \) and let's see, when \( x = 3 \), let's calculate the slope. Wait, maybe we can just read from the graph. The lower line: from \( x=-1,y=-2 \) to \( x = 3 \), how much does \( y \) change? Let's see the grid. Each unit is 1. The line goes from \( (-1,-2) \) to, say, \( (3,-4) \)? Wait, no, let's count the steps. From \( x=-1 \) to \( x = 3 \), that's 4 units to the right. The \( y \) - value: from \( y=-2 \) to \( y=-4 \)? Wait, no, the slope: \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points. The closed circle is at \( (-1,-2) \), and when \( x = 3 \), let's see the \( y \) - value. Looking at the graph, the line passes through \( (3,-4) \)? Wait, no, maybe I made a mistake. Wait, the lower line: when \( x = 0 \), what's \( y \)? If the line goes from \( (-1,-2) \) with a slope. Wait, maybe it's easier: for \( x = 3 \), since \( x=3\geq - 1 \), we use the lower line. Let's see the graph, the lower line (the non - horizontal line) at \( x = 3 \), the \( y \) - value is \( - 4 \)? Wait, no, let's re - check the graph.

Wait, maybe the upper line is \( y = 3 \) for \( x < - 1 \), the lower line is \( y=-2+\frac{-4 + 2}{3+1}(x + 1)\)? No, maybe a better way:

1. For \( x=-5 \): \( x=-5 < - 1 \), so we use the upper horizontal line. The upper horizontal line has \( y = 3 \) (since the open circle is at \( (-1,3) \) and it's a horizontal line, so \( y \) is constant 3 for \( x < - 1 \)). So \( y = 3 \).

2. For \( x=-1 \): \( x=-1 \) is the point where the two pieces meet. The upper piece has an open circle at \( x=-1 \), so we use the lower piece. The lower piece has a closed circle at \( x=-1 \). Looking at the graph, the closed circle is at \( (-1,-2) \)? Wait, no, the grid: the closed circle is at \( x=-1,y=-2 \)? Wait, the y - axis: the upper line is at \( y = 3 \), the lower line is at \( y=-2 \) when \( x=-1 \)? Wait, maybe I messed up the \( y \) - coordinate. Let's count the grid lines. The upper horizontal line: between \( y = 2 \) and \( y = 4 \), so \( y = 3 \). The lower line: the closed circle is at \( x=-1,y=-2 \) (since between \( y=-4 \) and \( y = 0 \), so \( y=-2 \) at \( x=-1 \)). Then, for the lower line, let's find the equation. Let's take two points: \( (-1,-2) \) and \( (3,-4) \). The slope \( m=\frac{-4-(-2)}{3 - (-1)}=\frac{-2}{4}=-\frac{1}{2}\). The equation is \( y - (-2)=-\frac{1}{2}(x - (-1))\), \( y + 2=-\frac{1}{2}(x + 1) \). When \( x = 3 \), \( y+2=-\frac{1}{2}(4)\), \( y+2=-2 \), \( y=-4 \).

So:

- When \( x=-5 \), \( y = 3 \) (since \( x=-5 < - 1 \), use the upper horizontal line).
- When \( x=-1 \), \( y=-2 \) (since \( x=-1 \) is in the domain of the lower line, which has a closed circle at \( x=-1,y=-2 \)).
- When \( x = 3 \), \( y=-4 \) (using the equation of the lower line or reading from the graph).

Answer:

When \( x=-5 \), \( y = \boldsymbol{3}\); when \( x=-1 \), \( y=\boldsymbol{-2}\); when \( x = 3 \), \( y=\boldsymbol{-4}\)