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To determine which functions have a [tex]$y$[/tex]-intercept of [tex]$(0, 5)$[/tex], we need to find the value of each function when [tex]$x = 0$[/tex]. Let's evaluate each function step by step:
### 1. [tex]\( f(x) = 2(b)^x + 5 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2(b)^0 + 5 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = 2 \cdot 1 + 5 = 2 + 5 = 7 \][/tex]
So, [tex]\( f(x) = 2(b)^x + 5 \)[/tex] does not have a [tex]$y$[/tex]-intercept of 5.
### 2. [tex]\( f(x) = -5(b)^x + 10 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -5(b)^0 + 10 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = -5 \cdot 1 + 10 = -5 + 10 = 5 \][/tex]
So, [tex]\( f(x) = -5(b)^x + 10 \)[/tex] does have a [tex]$y$[/tex]-intercept of 5.
### 3. [tex]\( f(x) = 7(b)^x - 2 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 7(b)^0 - 2 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = 7 \cdot 1 - 2 = 7 - 2 = 5 \][/tex]
So, [tex]\( f(x) = 7(b)^x - 2 \)[/tex] does have a [tex]$y$[/tex]-intercept of 5.
### 4. [tex]\( f(x) = -3(b)^x - 5 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -3(b)^0 - 5 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = -3 \cdot 1 - 5 = -3 - 5 = -8 \][/tex]
So, [tex]\( f(x) = -3(b)^x - 5 \)[/tex] does not have a [tex]$y$[/tex]-intercept of 5.
### 5. [tex]\( f(x) = 5(b)^x - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 5(b)^0 - 1 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = 5 \cdot 1 - 1 = 5 - 1 = 4 \][/tex]
So, [tex]\( f(x) = 5(b)^x - 1 \)[/tex] does not have a [tex]$y$[/tex]-intercept of 5.
### Conclusion
The functions that have a [tex]$y$[/tex]-intercept of [tex]$(0, 5)$[/tex] are:
[tex]\[ f(x) = -5(b)^x + 10 \][/tex]
[tex]\[ f(x) = 7(b)^x - 2 \][/tex]
So, the correct answers are:
2 and 3.
### 1. [tex]\( f(x) = 2(b)^x + 5 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2(b)^0 + 5 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = 2 \cdot 1 + 5 = 2 + 5 = 7 \][/tex]
So, [tex]\( f(x) = 2(b)^x + 5 \)[/tex] does not have a [tex]$y$[/tex]-intercept of 5.
### 2. [tex]\( f(x) = -5(b)^x + 10 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -5(b)^0 + 10 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = -5 \cdot 1 + 10 = -5 + 10 = 5 \][/tex]
So, [tex]\( f(x) = -5(b)^x + 10 \)[/tex] does have a [tex]$y$[/tex]-intercept of 5.
### 3. [tex]\( f(x) = 7(b)^x - 2 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 7(b)^0 - 2 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = 7 \cdot 1 - 2 = 7 - 2 = 5 \][/tex]
So, [tex]\( f(x) = 7(b)^x - 2 \)[/tex] does have a [tex]$y$[/tex]-intercept of 5.
### 4. [tex]\( f(x) = -3(b)^x - 5 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -3(b)^0 - 5 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = -3 \cdot 1 - 5 = -3 - 5 = -8 \][/tex]
So, [tex]\( f(x) = -3(b)^x - 5 \)[/tex] does not have a [tex]$y$[/tex]-intercept of 5.
### 5. [tex]\( f(x) = 5(b)^x - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 5(b)^0 - 1 \][/tex]
Since [tex]\( (b)^0 = 1 \)[/tex]:
[tex]\[ f(0) = 5 \cdot 1 - 1 = 5 - 1 = 4 \][/tex]
So, [tex]\( f(x) = 5(b)^x - 1 \)[/tex] does not have a [tex]$y$[/tex]-intercept of 5.
### Conclusion
The functions that have a [tex]$y$[/tex]-intercept of [tex]$(0, 5)$[/tex] are:
[tex]\[ f(x) = -5(b)^x + 10 \][/tex]
[tex]\[ f(x) = 7(b)^x - 2 \][/tex]
So, the correct answers are:
2 and 3.
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