Question

1. find the value of f(-4). y = f(x)

Explanation:

Step1: Identify the graph's segment for \( x = -4 \)

The left - hand segment of the piece - wise function \( y = f(x) \) (the increasing line) passes through \( (-9,0) \) and the vertex (let's assume the vertex is at \( x=-1 \), \( y = 6 \) from the graph). First, we find the equation of the line for \( x\leq - 1 \). The slope \( m\) of the line passing through \( (x_1,y_1)=(-9,0) \) and \( (x_2,y_2)=(-1,6) \) is calculated as \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{6 - 0}{-1-(-9)}=\frac{6}{8}=\frac{3}{4} \). Using the point - slope form \( y - y_1=m(x - x_1) \), with \( (x_1,y_1)=(-9,0) \) and \( m = \frac{3}{4} \), we have \( y-0=\frac{3}{4}(x + 9) \), so \( y=\frac{3}{4}x+\frac{27}{4} \).

Step2: Substitute \( x=-4 \) into the equation

Now, substitute \( x=-4 \) into \( y=\frac{3}{4}x+\frac{27}{4} \). We get \( y=\frac{3}{4}\times(-4)+\frac{27}{4}=\frac{-12 + 27}{4}=\frac{15}{4}=3.75 \)? Wait, maybe a better way: Looking at the graph, the left line (for \( x\leq - 1 \)): when \( x=-9 \), \( y = 0 \); when \( x=-1 \), \( y = 6 \). The change in \( x \) from \( - 9 \) to \( - 1 \) is \( 8 \) units, and the change in \( y \) is \( 6 \) units. The number of units from \( x=-9 \) to \( x = - 4 \) is \( (-4)-(-9)=5 \) units. Since the slope is \( \frac{6}{8}=\frac{3}{4} \), the change in \( y \) for a change of \( 5 \) in \( x \) is \( \frac{3}{4}\times5=\frac{15}{4} \)? No, wait, maybe we can use the two - point formula more simply. Alternatively, notice that the left line goes from \( (-9,0) \) to \( (-1,6) \). The equation can also be thought of as a linear function. Let's list the points: when \( x=-9 \), \( y = 0 \); \( x=-8 \), \( y=\frac{3}{4} \); \( x=-7 \), \( y=\frac{6}{4}=\frac{3}{2} \); \( x=-6 \), \( y=\frac{9}{4} \); \( x=-5 \), \( y=\frac{12}{4}=3 \); \( x=-4 \), \( y=\frac{15}{4}=3.75 \)? Wait, no, maybe I made a mistake. Wait, looking at the graph, the left - hand line: when \( x=-9 \), \( y = 0 \); when \( x=-1 \), \( y = 6 \). Let's count the grid. Each grid square is 1 unit. From \( x=-9 \) (where \( y = 0 \)) to \( x=-4 \), we move 5 units to the right. The slope is \( \frac{6}{8}=\frac{3}{4} \), so the rise is \( \frac{3}{4}\times5=\frac{15}{4}=3.75 \). But wait, maybe the vertex is at \( x=-1 \), \( y = 6 \), and the left line: when \( x=-9 \), \( y = 0 \), so the equation is \( y=\frac{6-0}{-1 - (-9)}(x + 9)=\frac{6}{8}(x + 9)=\frac{3}{4}(x + 9) \). When \( x=-4 \), \( y=\frac{3}{4}(-4 + 9)=\frac{3}{4}\times5=\frac{15}{4}=3.75 \)? But maybe the graph is such that the left line passes through \( (-9,0) \) and \( (-1,6) \), and when \( x=-4 \), we can also see from the graph (by counting the grid) that the \( y \) - value at \( x=-4 \) is 3? Wait, no, let's re - examine. Wait, the x - axis: from \( - 9 \) to \( - 1 \) is 8 units, and the y - axis from 0 to 6 is 6 units. So each unit in x (from - 9 to - 1) corresponds to \( \frac{6}{8}=\frac{3}{4} \) units in y. From \( x=-9 \) to \( x=-4 \), the number of x - units is \( (-4)-(-9)=5 \). So the y - value is \( 0+\frac{3}{4}\times5=\frac{15}{4}=3.75 \). But maybe the graph is drawn with integer coordinates. Wait, maybe I misread the graph. Let's look again. The left line: when \( x=-9 \), \( y = 0 \); when \( x=-5 \), \( y = 3 \); when \( x=-1 \), \( y = 6 \). Ah! So from \( x=-9 \) to \( x=-5 \), that's 4 units in x, and 3 units in y. So the slope is \( \frac{3}{4} \)? No, from \( x=-9 \) to \( x=-5 \), \( \Delta x=4 \), \( \Delta y = 3 \), so slope \( m=\frac{3}{4} \). Then from \( x=-9 \) to \( x=-4 \), \( \Delta x = 5 \), \( \Delta y=\frac{3}{4}\times5=\frac{15}{4}=3.75 \). But maybe the graph is such that at \( x=-4 \), the y - value is 3? Wait, no, let's check the points. If \( x=-9 \), \( y = 0 \); \( x=-8 \), \( y=\frac{3}{4} \); \( x=-7 \), \( y=\frac{6}{4}=\frac{3}{2} \); \( x=-6 \), \( y=\frac{9}{4} \); \( x=-5 \), \( y = 3 \); \( x=-4 \), \( y=\frac{15}{4}=3.75 \); \( x=-3 \), \( y=\frac{18}{4}=\frac{9}{2} \); \( x=-2 \), \( y=\frac{21}{4} \); \( x=-1 \), \( y = 6 \). So \( f(-4)=\frac{15}{4}=3.75 \) or \( 3\frac{3}{4} \). But maybe the graph is designed with a simpler slope. Wait, another way: the left line connects \( (-9,0) \) and \( (-1,6) \). The equation is \( y=\frac{6}{8}(x + 9)=\frac{3}{4}x+\frac{27}{4} \). When \( x=-4 \), \( y=\frac{3}{4}\times(-4)+\frac{27}{4}=\frac{-12 + 27}{4}=\frac{15}{4}=3.75 \).

Wait, maybe I made a mistake in the vertex. Let's look at the graph again. The graph has two segments: the left - hand segment (increasing) and the right - hand segment (decreasing). The right - hand segment passes through \( (0,3) \) and \( (4, - 1) \), slope \( \frac{-1 - 3}{4-0}=\frac{-4}{4}=-1 \), equation \( y=-x + 3 \). The left - hand segment: let's find two points. When \( x=-9 \), \( y = 0 \); when \( x=-1 \), \( y = 6 \). So the slope is \( \frac{6-0}{-1-(-9)}=\frac{6}{8}=\frac{3}{4} \), equation \( y=\frac{3}{4}(x + 9) \). When \( x=-4 \), substitute into the left - hand equation: \( y=\frac{3}{4}(-4 + 9)=\frac{3}{4}\times5=\frac{15}{4}=3.75 \).

Answer:

\( \frac{15}{4} \) (or \( 3.75 \))