Question
how does the graph of the function h(x)=4+5^x differ from the graph of g(x)=5^x
Answers
GPT-4o mini
To understand how the graph of the function \( h(x) = 4 + 5^x \) differs from the graph of \( g(x) = 5^x \), we need to analyze their components:
1. **Base Function**: The function \( g(x) = 5^x \) is an exponential function. Its characteristics include:
- It passes through the point \( (0, 1) \) since \( 5^0 = 1 \).
- As \( x \) approaches negative infinity, \( g(x) \) approaches 0. As \( x \) approaches positive infinity, \( g(x) \) goes to positive infinity.
- The graph rises steeply for positive \( x \) values.
2. **Transformation**: The function \( h(x) = 4 + 5^x \) is a vertical transformation of \( g(x) \). Here's how:
- The term \( +4 \) shifts the entire graph of \( g(x) \) upwards by 4 units.
- Therefore, the key points on the graph will change. For example, \( h(0) = 4 + 5^0 = 4 + 1 = 5 \). Thus, instead of passing through \( (0, 1) \), it passes through \( (0, 5) \).
- Similarly, as \( x \to -\infty \), \( h(x) \) approaches 4 instead of 0.
3. **End Behavior**:
- For \( g(x) \), as \( x \to -\infty\), \( g(x) \to 0 \), whereas for \( h(x) \), as \( x \to -\infty\), \( h(x) \to 4 \).
- For both functions, as \( x \to \infty\), the behavior is the same since they both involve \( 5^x \), so both approach \( \infty \).
In summary:
- The graph of \( h(x) = 4 + 5^x \) is identical to the graph of \( g(x) = 5^x \), but it is shifted upward by 4 units. This shift affects the y-values but does not change the overall shape of the exponential curve.
1. **Base Function**: The function \( g(x) = 5^x \) is an exponential function. Its characteristics include:
- It passes through the point \( (0, 1) \) since \( 5^0 = 1 \).
- As \( x \) approaches negative infinity, \( g(x) \) approaches 0. As \( x \) approaches positive infinity, \( g(x) \) goes to positive infinity.
- The graph rises steeply for positive \( x \) values.
2. **Transformation**: The function \( h(x) = 4 + 5^x \) is a vertical transformation of \( g(x) \). Here's how:
- The term \( +4 \) shifts the entire graph of \( g(x) \) upwards by 4 units.
- Therefore, the key points on the graph will change. For example, \( h(0) = 4 + 5^0 = 4 + 1 = 5 \). Thus, instead of passing through \( (0, 1) \), it passes through \( (0, 5) \).
- Similarly, as \( x \to -\infty \), \( h(x) \) approaches 4 instead of 0.
3. **End Behavior**:
- For \( g(x) \), as \( x \to -\infty\), \( g(x) \to 0 \), whereas for \( h(x) \), as \( x \to -\infty\), \( h(x) \to 4 \).
- For both functions, as \( x \to \infty\), the behavior is the same since they both involve \( 5^x \), so both approach \( \infty \).
In summary:
- The graph of \( h(x) = 4 + 5^x \) is identical to the graph of \( g(x) = 5^x \), but it is shifted upward by 4 units. This shift affects the y-values but does not change the overall shape of the exponential curve.