Question

graph the piecewise - defined function.
$f(x)=\begin{cases}x - 1&\text{if }xleq - 3\\-4&\text{if }x > - 3end{cases}$

choose the correct graph
\\(\bigcirc\\) a.
\\(\bigcirc\\) b.
\\(\bigcirc\\) c.
\\(\bigcirc\\) d.

Explanation:

Step1: Analyze the first piece \( f(x) = x - 1 \) for \( x \leq -3 \)

This is a linear function with slope \( 1 \) and y-intercept \( -1 \). We can find a point on this line by plugging \( x = -3 \): \( f(-3)=-3 - 1=-4 \). So the point \( (-3, -4) \) is on this part, and since the slope is positive, as \( x \) decreases (because \( x \leq -3 \)), \( y \) decreases.

Step2: Analyze the second piece \( f(x)=-4 \) for \( x > -3 \)

This is a horizontal line \( y = -4 \) for all \( x \) values greater than \( -3 \). At \( x=-3 \), the first piece gives \( y=-4 \), and the second piece starts at \( x > -3 \), so there is a closed dot at \( (-3, -4) \) for the first piece and an open dot? Wait, no: for \( x \leq -3 \), the first piece includes \( x=-3 \), so the point \( (-3, -4) \) is included there. For \( x > -3 \), the second piece is \( y=-4 \), so the graph should have a line with slope 1 (from \( f(x)=x - 1 \)) for \( x \leq -3 \) (passing through \( (-3, -4) \) and decreasing as \( x \) goes left), and a horizontal line \( y=-4 \) for \( x > -3 \).

Now let's check the options:

- Option A: The line for \( x \leq -3 \) has negative slope (correct, since \( f(x)=x - 1 \) has slope 1? Wait no, slope of \( x - 1 \) is 1, positive. Wait, I made a mistake: \( f(x)=x - 1 \) has slope 1, so as \( x \) increases, \( y \) increases. But for \( x \leq -3 \), so when \( x=-3 \), \( y=-4 \); when \( x=-4 \), \( y=-5 \), etc. So the line for \( x \leq -3 \) should have positive slope, going from \( (-3, -4) \) down to the left (since \( x \) is less than -3, so moving left from \( x=-3 \), \( x \) decreases, \( y = x - 1 \) decreases, so the line goes from \( (-3, -4) \) with slope 1 (so for each unit left, \( y \) decreases by 1). Then for \( x > -3 \), it's a horizontal line \( y=-4 \).

Wait let's re-express:

For \( f(x)=x - 1 \), when \( x=-3 \), \( y=-4 \); \( x=-4 \), \( y=-5 \); \( x=-5 \), \( y=-6 \), etc. So the line has slope 1, so it's a line with slope 1, passing through \( (-3, -4) \), and for \( x \leq -3 \), we plot that. Then for \( x > -3 \), \( y=-4 \), so a horizontal line at \( y=-4 \) starting from \( x > -3 \) (so open dot at \( (-3, -4) \)? No, wait: the first piece is \( x \leq -3 \), so \( x=-3 \) is included in the first piece, so the point \( (-3, -4) \) is on the first piece. The second piece is \( x > -3 \), so at \( x=-3 \), the second piece does not include \( x=-3 \), but the first piece does. Wait, no: the function is defined as \( x \leq -3 \) for the first piece, so \( x=-3 \) is in the first piece, and \( x > -3 \) for the second. So the graph should have a closed dot at \( (-3, -4) \) for the first piece, and then a horizontal line \( y=-4 \) for \( x > -3 \).

Wait, let's recast:

- First piece: \( f(x)=x - 1 \), \( x \leq -3 \). So this is a linear function with slope 1, so it's a line going up to the right (since slope is positive). But since \( x \leq -3 \), the domain for this piece is \( x \leq -3 \), so we take the part of the line \( y = x - 1 \) where \( x \leq -3 \). So when \( x=-3 \), \( y=-4 \); when \( x=-4 \), \( y=-5 \); when \( x=-5 \), \( y=-6 \), etc. So the line segment (or ray) for \( x \leq -3 \) starts at \( (-3, -4) \) and goes to the left (decreasing \( x \)) with slope 1 (so decreasing \( y \)).

Second piece: \( y=-4 \) for \( x > -3 \), so a horizontal line starting just to the right of \( x=-3 \) (open dot at \( (-3, -4) \) for the second piece? No, the first piece includes \( x=-3 \), so the first piece has a closed dot at \( (-3, -4) \), and the second piece has an open dot at \( (-3, -4) \) (since \( x > -3 \) does not include \( x=-3 \))? Wait, no: the function's value at \( x=-3 \) is given by the first piece: \( f(-3)=-3 - 1=-4 \). The second piece is for \( x > -3 \), so at \( x=-3 \), it's defined by the first piece. So the graph should have a closed dot at \( (-3, -4) \) from the first piece, and then a horizontal line \( y=-4 \) for \( x > -3 \) (so the point \( (-3, -4) \) is included in the first piece, and the second piece starts at \( x > -3 \), so the horizontal line starts with an open dot? No, no: the function is continuous at \( x=-3 \) because \( f(-3)=-4 \) (from first piece) and the limit as \( x \) approaches -3 from the right is \( -4 \) (since \( f(x)=-4 \) for \( x > -3 \)), so the graph should have a closed dot at \( (-3, -4) \) for both? Wait, no: the first piece is \( x \leq -3 \), so includes \( x=-3 \), the second is \( x > -3 \), so does not include \( x=-3 \). But since \( f(-3)=-4 \) (from first piece) and the second piece is \( y=-4 \) for \( x > -3 \), the graph will have a closed dot at \( (-3, -4) \) (from first piece) and the horizontal line \( y=-4 \) for \( x > -3 \) (so the point \( (-3, -4) \) is included, and the horizontal line continues to the right). Wait, maybe I confused the open/closed dots. Let's re-express:

- For \( f(x)=x - 1 \), \( x \leq -3 \): this is a line with slope 1, passing through \( (-3, -4) \). So when \( x=-3 \), \( y=-4 \) (closed dot). For \( x < -3 \), say \( x=-4 \), \( y=-5 \); \( x=-5 \), \( y=-6 \), etc. So this line goes from \( (-3, -4) \) with slope 1 (so for each \( x \) decrease by 1, \( y \) decreases by 1), so it's a line going down to the left from \( (-3, -4) \).

- For \( f(x)=-4 \), \( x > -3 \): this is a horizontal line \( y=-4 \) for all \( x > -3 \). So at \( x=-3 \), the first piece gives \( y=-4 \), and for \( x > -3 \), \( y=-4 \), so the graph is a line with slope 1 (from \( f(x)=x - 1 \)) for \( x \leq -3 \) (passing through \( (-3, -4) \) and going left with slope 1) and a horizontal line \( y=-4 \) for \( x > -3 \) (starting at \( x > -3 \), so the point \( (-3, -4) \) is included in the first piece, and the horizontal line continues to the right).

Now let's look at the options:

- Option B: The left part (for \( x \leq -3 \)) is a line with slope -1? No, wait the line in option B: when \( x=-3 \), \( y=-4 \), and as \( x \) decreases (goes left), \( y \) increases? No, that would be slope -1. Wait, no, \( f(x)=x - 1 \) has slope 1, so when \( x \) decreases by 1 (from -3 to -4), \( y \) decreases by 1 (from -4 to -5). So the line should go from \( (-3, -4) \) down to the left (slope 1). Let's check the graphs:

Wait maybe I made a mistake in slope. Wait \( f(x)=x - 1 \): slope is 1, so it's a line going up to the right. But for \( x \leq -3 \), so the domain is \( x \leq -3 \), so the graph of this part is the ray of \( y = x - 1 \) where \( x \leq -3 \). So when \( x=-3 \), \( y=-4 \); when \( x=-4 \), \( y=-5 \); \( x=-5 \), \( y=-6 \), etc. So the ray starts at \( (-3, -4) \) and goes to the left (decreasing \( x \)) with slope 1 (so decreasing \( y \)), which is a line with positive slope (since slope is rise over run, run is negative (left), rise is negative (down), so slope is positive).

Then the second part is \( y=-4 \) for \( x > -3 \), so a horizontal line to the right of \( x=-3 \).

Now let's check the options:

- Option B: The left part (x <= -3) is a line with slope -1? No, wait the line in option B: when x=-3, y=-4, and as x decreases (left), y increases (so slope -1), which is wrong. Wait no, maybe I mixed up the direction. Wait, slope 1: (y2 - y1)/(x2 - x1) = 1. If x2 = x1 - 1 (decrease x by 1), then y2 = y1 - 1 (decrease y by 1). So the line goes from (-3, -4) to (-4, -5), which is down and left, so the slope is (-5 - (-4))/(-4 - (-3)) = (-1)/(-1) = 1, correct. So the line has positive slope, going down to the left.

Now the horizontal line for x > -3 is y=-4, so to the right of x=-3, it's a horizontal line at y=-4.

Looking at the options, option B: the left part is a line with slope -1? No, wait maybe the options are:

Wait the user's options:

Option B's graph: left part (x <= -3) is a line with slope -1? No, let's re-express:

Wait the correct graph should have:

- For x <= -3: a line with slope 1 (so as x increases, y increases), but since x <= -3, the line is from x=-3 (y=-4) going to the left (x decreasing) with slope 1 (y decreasing).

- For x > -3: a horizontal line y=-4.

Now, let's check the options:

Option B: The left line (x <= -3) has a closed dot at (-3, -4), and as x decreases, y decreases (slope 1), and the right part is horizontal y=-4. Wait maybe I was wrong earlier, let's check the slope again. \( f(x)=x - 1 \) has slope 1, so it's a line that goes up 1 for each 1 to the right. So for x <= -3, the line is the part of y = x - 1 where x is less than or equal to -3. So when x = -3, y = -4; when x = -2, y = -3, but x=-2 is greater than -3, so that's not in the first piece. Wait, no! Wait x <= -3, so x can be -3, -4, -5, etc. So when x = -3, y = -4; x = -4, y = -5; x = -5, y = -6, etc. So the line for x <= -3 is the segment (ray) from (-3, -4) going to the left (x decreasing) with slope 1 (so y decreasing). Then for x > -3, y = -4, so x = -2, y = -4; x = -1, y = -4; x = 0, y = -4, etc. So the graph is a line with slope 1 (going left and down) for x <= -3, and a horizontal line y=-4 for x > -3.

Now looking at the options, option B has a line on the left (x <= -3) with slope -1? No, wait maybe the options are:

Wait the user's option B: the left part is a line that goes from (-3, -4) up to the left (slope -1), which is wrong. Wait no, maybe I made a mistake in the function. Wait \( f(x)=x - 1 \) when x <= -3: so when x = -3, y = -4; x = -4, y = -5; x = -5, y = -6. So the line has a positive slope (1), so it's a line that, as x decreases (moves left), y decreases (moves down). So the line goes from (-3, -4) down to the left. Then the right part is horizontal y=-4.

Now let's check the options:

- Option B: The left line (x <= -3) is a line with slope -1 (goes up to the left), which is wrong.

- Option D: Wait no, let's think again. Wait maybe the correct option is B? Wait no, let's take a point: for x = -4 (which is <= -3), f(-4) = -4 - 1 = -5. So the point (-4, -5) should be on the left part. For x = 0 (which is > -3), f(0) = -4. So the point (0, -4) should be on the right part.

Now let's look at the graphs:

Option B: The left part has a point at (-3, -4), and as x decreases (goes left), y decreases (so (-4, -5) is on it), and the right part is horizontal at y=-4. That matches.

Option A: The left part is decreasing with slope -1 (goes up to the left), which is wrong.

Option C: The left part is increasing with slope 1 (goes up to the right), but x <= -3, so that would be for x >= -3, which is wrong.

Option D: The left part is increasing with slope 1, starting at (-3, -4) and going up to the right, but that's for x >= -3, which is wrong.

So the correct option is B.

Answer:

B