what are the features of function g if ( g(x) = -f(x) - 1 )?
( y )-intercept at ( (0, -1) )
range of ( (-1, infty) )
domain of ( (0, infty) )
( x )-intercept at ( left( \frac{1}{2}, 0
ight) )
vertical asymptote of ( x = 0 )

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Accepted Answer

To solve this, we analyze the transformation $ g(x)=-f(x)-1 $ and the original graph of $ f(x) $:

### Step 1: Analyze the original function $ f(x) $
From the graph, $ f(x) $ appears to be an exponential or logarithmic - like curve. Let's assume the original $ f(x) $ has a $ y $ - intercept at $ (0,0) $ (since it crosses the $ y $ - axis at $ x = 0,y = 0 $) and domain $ (0,\infty) $, range $ (- \infty,0) $, $ x $ - intercept (if any) and vertical asymptote $ x = 0 $.

### Step 2: Analyze the transformation for $ g(x)=-f(x)-1 $
- **Y - intercept**:
For the $ y $ - intercept, we set $ x = 0 $. Then $ g(0)=-f(0)-1 $. If the original $ f(0) = 0 $ (from the graph, the curve passes through $ (0,0) $), then $ g(0)=-0 - 1=-1 $. So the $ y $ - intercept of $ g(x) $ is at $ (0,-1) $.
- **Range**:
The range of $ f(x) $ (from the graph) seems to be $ (- \infty,0) $. When we apply the transformation $ y=-f(x)-1 $, first, the transformation $ y = - f(x) $ reflects $ f(x) $ over the $ x $ - axis. So the range of $ -f(x) $ is $ (0,\infty) $. Then, the transformation $ y=-f(x)-1 $ shifts the graph down by 1 unit. So the range of $ g(x)=-f(x)-1 $ is $ (-1,\infty) $.
- **Domain**:
The domain of a function is affected by vertical transformations (reflections and vertical shifts) in the same way as the original function. If the domain of $ f(x) $ is $ (0,\infty) $, the domain of $ g(x)=-f(x)-1 $ is also $ (0,\infty) $ (since vertical transformations do not affect the domain).
- **X - intercept**:
To find the $ x $ - intercept, we set $ g(x) = 0 $, so $ 0=-f(x)-1\Rightarrow f(x)=- 1 $. There is no information from the original graph to suggest that $ f(x)=-1 $ at $ x=\frac{1}{2} $, so this option is incorrect.
- **Vertical Asymptote**:
Vertical asymptotes are affected by horizontal transformations. Since $ g(x) $ is a vertical transformation of $ f(x) $, if the original $ f(x) $ has a vertical asymptote $ x = 0 $, the vertical asymptote of $ g(x) $ is also $ x = 0 $. But we are checking the options one by one.

Now let's check each option:
- Option 1: $ y $ - intercept at $ (0,-1) $: As we calculated above, when $ x = 0 $, $ g(0)=-f(0)-1 $. If $ f(0) = 0 $ (from the graph of $ f(x) $ passing through $ (0,0) $), then $ g(0)=-1 $. So this is correct.
- Option 2: Range of $ (-1,\infty) $: The range of $ f(x) $ is $ (-\infty,0) $. For $ y=-f(x)-1 $, when we reflect $ f(x) $ over the $ x $ - axis, the range of $ -f(x) $ is $ (0,\infty) $. Then shifting down by 1 unit, the range of $ -f(x)-1 $ is $ (-1,\infty) $. This is correct.
- Option 3: Domain of $ (0,\infty) $: Vertical transformations (reflection over $ x $ - axis and vertical shift) do not change the domain. If the domain of $ f(x) $ is $ (0,\infty) $, the domain of $ g(x) $ is also $ (0,\infty) $. This is correct.
- Option 4: $ x $ - intercept at $ (\frac{1}{2},0) $: To find the $ x $ - intercept, we set $ g(x) = 0\Rightarrow -f(x)-1=0\Rightarrow f(x)=-1 $. There is no evidence from the graph of $ f(x) $ that $ f(\frac{1}{2})=-1 $, so this is incorrect.
- Option 5: Vertical asymptote of $ x = 0 $: Vertical asymptotes are determined by the values of $ x $ where the function is undefined. Vertical transformations (reflection and vertical shift) do not change the vertical asymptote. If $ f(x) $ has a vertical asymptote at $ x = 0 $, $ g(x) $ also has a vertical asymptote at $ x = 0 $. This is correct. But we are to choose the features of $ g(x) $. Since the question is about the features of $ g(x) $ and we are to find which is a feature, the $ y $ - intercept at $ (0,-1) $, range $ (-1,\infty) $, domain $ (0,\infty) $ and vertical asymptote $ x = 0 $ are all features. But if we consider the most direct one from the transformation of the $ y $ - intercept:

If we have to choose one of the options as a feature (assuming the question is to select the correct feature from the given list), the $ y $ - intercept at $ (0,-1) $ is a correct feature. Also, the range $ (-1,\infty) $, domain $ (0,\infty) $ and vertical asymptote $ x = 0 $ are correct. But if we take the $ y $ - intercept:

The answer for the $ y $ - intercept feature: $ y $ - intercept at $ (0,-1) $

If we consider the range: The range of $ g(x) $ is $ (-1,\infty) $

If we consider the domain: The domain of $ g(x) $ is $ (0,\infty) $

If we consider the vertical asymptote: The vertical asymptote is $ x = 0 $

If we are to choose the correct option from the given list (the yellow - highlighted option is $ y $ - intercept at $ (0,-1) $):

The correct option is the one with $ y $ - intercept at $ (0,-1) $, range of $ (-1,\infty) $, domain of $ (0,\infty) $ and vertical asymptote $ x = 0 $. But if we take the first option (yellow - highlighted) $ y $ - intercept at $ (0,-1) $ is correct.

So, for the $ y $ - intercept feature:
# Explanation:
## Step 1: Find $ g(0) $
Set $ x = 0 $ in $ g(x)=-f(x)-1 $. If $ f(0) = 0 $ (from the graph of $ f(x) $), then $ g(0)=-f(0)-1=-0 - 1=-1 $.
## Step 2: Determine the $ y $ - intercept
The $ y $ - intercept occurs at $ x = 0 $. So the $ y $ - intercept of $ g(x) $ is $ (0,-1) $.
# Answer:
$ y $ - intercept at $ (0,-1) $ (along with range $ (-1,\infty) $, domain $ (0,\infty) $ and vertical asymptote $ x = 0 $ as correct features)

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QUESTION IMAGE

Question

Explanation:

Step 1: Analyze the original function \( f(x) \)

Step 2: Analyze the transformation for \( g(x)=-f(x)-1 \)

Step 1: Find \( g(0) \)

Step 2: Determine the \( y \) - intercept

Answer: