Question

1. use the graph $y = f(x)$ below.

what is the graph of $g(x) = \frac{1}{2}f(x) - 6$?

(images of graphs for options a, b, c, d are shown)

Explanation:

Step1: Analyze Vertical Stretch/Compression

The function \( g(x)=\frac{1}{2}f(x)-6 \) involves a vertical compression by a factor of \( \frac{1}{2} \) (since we multiply \( f(x) \) by \( \frac{1}{2} \)) and a vertical shift down by 6 units. First, consider key points of \( f(x) \). For example, the y - intercept of \( f(x) \) is at \( (0,8) \), and the x - intercept is at \( (8,0) \), and the peak of the left - most part (let's say at \( x = - 8 \)) has a y - value of 16.

Step2: Apply Vertical Compression

For the y - intercept \( (0,8) \): After vertical compression by \( \frac{1}{2} \), the y - value becomes \( \frac{1}{2}\times8 = 4 \). Then, after shifting down by 6 units, the y - value becomes \( 4-6=-2 \)? Wait, no, wait. Wait, the order of operations: first apply the vertical stretch/compression, then the vertical shift. Wait, \( g(x)=\frac{1}{2}f(x)-6 \), so we first compute \( \frac{1}{2}f(x) \), then subtract 6.

For the point \( (0,8) \) on \( f(x) \): \( \frac{1}{2}f(0)=\frac{1}{2}\times8 = 4 \), then \( g(0)=\frac{1}{2}f(0)-6=4 - 6=-2 \)? Wait, maybe I made a mistake. Wait, let's take another point. The peak at \( x=-8 \): \( f(-8) = 16 \), so \( \frac{1}{2}f(-8)=\frac{1}{2}\times16 = 8 \), then \( g(-8)=8 - 6 = 2 \)? Wait, no, maybe I should look at the shape. The graph of \( f(x) \) has a linear part from \( x=-4 \) to \( x = 8 \) with slope \( m=\frac{0 - 8}{8-0}=\frac{-8}{8}=-1 \). After vertical compression by \( \frac{1}{2} \), the slope of the linear part becomes \( \frac{1}{2}\times(-1)=-\frac{1}{2} \), and then we shift down by 6 units.

Wait, let's check the x - intercept of \( f(x) \) is \( (8,0) \). For \( g(x) \), set \( g(x) = 0 \): \( 0=\frac{1}{2}f(x)-6\Rightarrow\frac{1}{2}f(x)=6\Rightarrow f(x)=12 \). Wait, maybe a better approach is to analyze the transformation step - by - step.

Vertical compression by \( \frac{1}{2} \): All y - coordinates of \( f(x) \) are multiplied by \( \frac{1}{2} \). Then vertical shift down by 6 units: all y - coordinates are decreased by 6.

Take the y - intercept of \( f(x) \): \( (0,8) \). After compression: \( (0,4) \). After shift down: \( (0,4 - 6)=(0,-2) \).

Take the x - intercept of \( f(x) \): \( (8,0) \). After compression: \( (8,0) \) (since \( \frac{1}{2}\times0 = 0 \)). After shift down: \( (8,0 - 6)=(8,-6) \).

Take the peak at \( x=-8 \): \( f(-8)=16 \). After compression: \( ( - 8,8) \). After shift down: \( (-8,8 - 6)=( - 8,2) \).

Now, let's look at the options. Option b: Let's check the graph of option b. The graph of \( g(x) \) in option b has a linear part with slope \( -\frac{1}{2} \) (since the original slope of the linear part of \( f(x) \) from \( x=-4 \) to \( x = 8 \) is \( - 1 \), after compression by \( \frac{1}{2} \), slope is \( - \frac{1}{2} \)) and the vertical shift down by 6. The key is that the vertical compression makes the graph "shorter" vertically, and the shift down moves it down.

Looking at the options, option b seems to have the correct transformation. The original \( f(x) \) has a linear segment from \( (0,8) \) to \( (8,0) \) (slope - 1). After compression by \( \frac{1}{2} \), the slope becomes \( - \frac{1}{2} \), and after shifting down by 6, the line \( g(x) \) will have a y - intercept lower and the x - intercept shifted? Wait, maybe another way: the equation of the linear part of \( f(x) \) (from \( x=-4 \) to \( x = 8 \)) is \( y=-x + 8 \) (since when \( x = 0 \), \( y = 8 \); when \( x = 8 \), \( y=0 \)). Then \( g(x)=\frac{1}{2}(-x + 8)-6=-\frac{1}{2}x+4 - 6=-\frac{1}{2}x-2 \). The y - intercept of \( g(x) \) is \( (0,-2) \) and the x - intercept is when \( -\frac{1}{2}x-2 = 0\Rightarrow-\frac{1}{2}x=2\Rightarrow x=-4 \)? Wait, no, that can't be. Wait, maybe my equation of the linear part of \( f(x) \) is wrong. Wait, from \( x=-4 \) to \( x = 8 \), when \( x=-4 \), what's the y - value? Looking at the graph, at \( x=-4 \), the y - value is 16? Wait, no, the graph of \( f(x) \): from \( x=-16 \) to \( x=-12 \), it's a line, then from \( x=-12 \) to \( x=-8 \), it goes down, then from \( x=-8 \) to \( x=-4 \), it goes up to 16, then from \( x=-4 \) to \( x = 8 \), it goes down to 0. So the linear part from \( x=-4 \) to \( x = 8 \): when \( x=-4 \), \( y = 16 \); when \( x = 8 \), \( y = 0 \). So the slope \( m=\frac{0 - 16}{8-(-4)}=\frac{-16}{12}=-\frac{4}{3} \)? Wait, I made a mistake earlier. Oh no, I misidentified the points. Let's correctly identify the linear segment from \( x=-4 \) to \( x = 8 \). At \( x=-4 \), \( y = 16 \); at \( x = 0 \), \( y = 8 \); at \( x = 8 \), \( y = 0 \). So the slope between \( x=-4 \) and \( x = 8 \) is \( \frac{0 - 16}{8-(-4)}=\frac{-16}{12}=-\frac{4}{3} \)? No, between \( x = 0 \) and \( x = 8 \): \( \frac{0 - 8}{8-0}=-1 \), between \( x=-4 \) and \( x = 0 \): \( \frac{8 - 16}{0-(-4)}=\frac{-8}{4}=-2 \). So the function \( f(x) \) has different linear segments.

But maybe a better approach is to look at the transformation of the graph. The function \( g(x)=\frac{1}{2}f(x)-6 \) is a vertical compression (by 1/2) and vertical shift down (by 6). So the graph of \( g(x) \) should be "shorter" than \( f(x) \) vertically and shifted down. Looking at the options, option b has a graph that is a compressed and shifted version of \( f(x) \). The other options: option a has a steeper slope (wrong, since we have a compression, slope should be less steep in magnitude), option c has a horizontal shift (no, the transformation is vertical, not horizontal), option d has a different shape. So the correct graph is option b.

Answer:

b