Answer :
Let's analyze the function [tex]\( g \)[/tex] step by step:
### Step 1: Define the function
The function [tex]\( g \)[/tex] is defined piecewise as:
[tex]\[ g(x) = \begin{cases} \left(\frac{3}{4}\right)^x & \text{for } x < 0 \\ -x^2 & \text{for } x \geq 0 \end{cases} \][/tex]
### Step 2: Find intercepts
- [tex]\( x \)[/tex]-intercepts:
- For [tex]\( x \geq 0 \)[/tex], we set [tex]\( g(x) = -x^2 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[ -x^2 = 0 \][/tex]
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Therefore, there is one [tex]\( x \)[/tex]-intercept at [tex]\( x = 0 \)[/tex].
- [tex]\( y \)[/tex]-intercepts:
- The [tex]\( y \)[/tex]-intercept is the value of [tex]\( g \)[/tex] when [tex]\( x = 0 \)[/tex]. For [tex]\( x = 0 \)[/tex], we use the part of the piecewise function that applies to [tex]\( x \geq 0 \)[/tex]:
[tex]\[ g(0) = -0^2 = 0 \][/tex]
Therefore, there is one [tex]\( y \)[/tex]-intercept at [tex]\( y = 0 \)[/tex].
### Step 3: Determine where the function is decreasing
- For [tex]\( x < 0 \)[/tex], [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex]:
- [tex]\( \left( \frac{3}{4} \right)^x \)[/tex] decreases as [tex]\( x \to -\infty \)[/tex] and increases as [tex]\( x \to 0 \)[/tex], but we consider it increasing when [tex]\( x < 0 \)[/tex].
- For [tex]\( x \geq 0 \)[/tex], [tex]\( g(x) = -x^2 \)[/tex]:
- [tex]\( -x^2 \)[/tex] is a decreasing function for [tex]\( x \geq 0 \)[/tex].
Combining these observations, [tex]\( g(x) \)[/tex] is decreasing for [tex]\( x \geq 0 \)[/tex].
### Step 4: Continuity
- To check continuity, the function must be continuous at the point where the pieces of the function meet, which is [tex]\( x = 0 \)[/tex].
- As [tex]\( x \to 0^- \)[/tex], the value of [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex] approaches [tex]\( 1 \)[/tex].
- As [tex]\( x \to 0^+ \)[/tex], the value of [tex]\( g(x) = -x^2 \)[/tex] goes to [tex]\( 0 \)[/tex].
Since there is a jump from 1 to 0 at [tex]\( x = 0 \)[/tex], the function [tex]\( g \)[/tex] is not continuous at [tex]\( x = 0 \)[/tex].
### Step 5: Conclusion
Based on the above analysis:
- The function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- The function [tex]\( g \)[/tex] is sometimes decreasing.
- The function [tex]\( g \)[/tex] is not continuous.
### Final Result:
So, filling in the blanks we get:
- Function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- Function [tex]\( g \)[/tex] is sometimes decreasing.
- Function [tex]\( g \)[/tex] not continuous.
### Step 1: Define the function
The function [tex]\( g \)[/tex] is defined piecewise as:
[tex]\[ g(x) = \begin{cases} \left(\frac{3}{4}\right)^x & \text{for } x < 0 \\ -x^2 & \text{for } x \geq 0 \end{cases} \][/tex]
### Step 2: Find intercepts
- [tex]\( x \)[/tex]-intercepts:
- For [tex]\( x \geq 0 \)[/tex], we set [tex]\( g(x) = -x^2 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[ -x^2 = 0 \][/tex]
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Therefore, there is one [tex]\( x \)[/tex]-intercept at [tex]\( x = 0 \)[/tex].
- [tex]\( y \)[/tex]-intercepts:
- The [tex]\( y \)[/tex]-intercept is the value of [tex]\( g \)[/tex] when [tex]\( x = 0 \)[/tex]. For [tex]\( x = 0 \)[/tex], we use the part of the piecewise function that applies to [tex]\( x \geq 0 \)[/tex]:
[tex]\[ g(0) = -0^2 = 0 \][/tex]
Therefore, there is one [tex]\( y \)[/tex]-intercept at [tex]\( y = 0 \)[/tex].
### Step 3: Determine where the function is decreasing
- For [tex]\( x < 0 \)[/tex], [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex]:
- [tex]\( \left( \frac{3}{4} \right)^x \)[/tex] decreases as [tex]\( x \to -\infty \)[/tex] and increases as [tex]\( x \to 0 \)[/tex], but we consider it increasing when [tex]\( x < 0 \)[/tex].
- For [tex]\( x \geq 0 \)[/tex], [tex]\( g(x) = -x^2 \)[/tex]:
- [tex]\( -x^2 \)[/tex] is a decreasing function for [tex]\( x \geq 0 \)[/tex].
Combining these observations, [tex]\( g(x) \)[/tex] is decreasing for [tex]\( x \geq 0 \)[/tex].
### Step 4: Continuity
- To check continuity, the function must be continuous at the point where the pieces of the function meet, which is [tex]\( x = 0 \)[/tex].
- As [tex]\( x \to 0^- \)[/tex], the value of [tex]\( g(x) = \left( \frac{3}{4} \right)^x \)[/tex] approaches [tex]\( 1 \)[/tex].
- As [tex]\( x \to 0^+ \)[/tex], the value of [tex]\( g(x) = -x^2 \)[/tex] goes to [tex]\( 0 \)[/tex].
Since there is a jump from 1 to 0 at [tex]\( x = 0 \)[/tex], the function [tex]\( g \)[/tex] is not continuous at [tex]\( x = 0 \)[/tex].
### Step 5: Conclusion
Based on the above analysis:
- The function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- The function [tex]\( g \)[/tex] is sometimes decreasing.
- The function [tex]\( g \)[/tex] is not continuous.
### Final Result:
So, filling in the blanks we get:
- Function [tex]\( g \)[/tex] has 1 [tex]\( x \)[/tex]-intercept(s) and 1 [tex]\( y \)[/tex]-intercept(s).
- Function [tex]\( g \)[/tex] is sometimes decreasing.
- Function [tex]\( g \)[/tex] not continuous.
