Answer
- The \( y \)-intercept of \( h \) is at \( (0,\ 5) \).
- \( h \) does not have an \( x \)-intercept because \( h(x) \) is always greater than 0.
- The graph of \( h \) has a horizontal asymptote at \( y = 1 \) and passes through points like \( (1,\ 3) \) and \( (-1,\ 9) \).
- Function \( g(x) = 4 \cdot (2^{-x} + 2) \) is obtained by shifting \( h(x) \) vertically upward by 7 units.
Solution
Let's analyze each part of the problem step by step for the function \( h(x) = 4 \cdot 2^{-x} + 1 \).
---
### **5.2.1 Determine the coordinates of the \( y \)-intercept of \( h \).**
**Solution:**
The \( y \)-intercept of a function occurs where \( x = 0 \).
\[
h(0) = 4 \cdot 2^{-0} + 1 = 4 \cdot 1 + 1 = 5
\]
**Answer:**
The \( y \)-intercept of \( h \) is at the point \( (0,\ 5) \).
---
### **5.2.2 Explain why \( h \) does not have an \( x \)-intercept.**
**Solution:**
An \( x \)-intercept occurs where \( h(x) = 0 \).
\[
0 = 4 \cdot 2^{-x} + 1 \\
4 \cdot 2^{-x} = -1 \\
2^{-x} = -\frac{1}{4}
\]
However, \( 2^{-x} = \frac{1}{2^x} \) is always positive for all real values of \( x \). Therefore, \( 2^{-x} \) can never be negative, making \( 4 \cdot 2^{-x} + 1 \) always greater than 0.
**Answer:**
Because \( h(x) = 4 \cdot 2^{-x} + 1 \) is always greater than zero for all real \( x \), the graph never crosses the \( x \)-axis. Hence, \( h \) does not have an \( x \)-intercept.
---
### **5.2.3 Draw a sketch graph of \( h \), clearly showing all asymptotes, intercepts with the axes and at least one other point on \( h \).**
**Solution:**
While a visual sketch cannot be drawn here, here's a detailed description of the graph:
1. **Y-Intercept:**
- As calculated in 5.2.1, the \( y \)-intercept is at \( (0,\ 5) \).
2. **Asymptote:**
- **Horizontal Asymptote:** \( y = 1 \)
- As \( x \to \infty \), \( 2^{-x} \to 0 \) so \( h(x) \to 1 \).
3. **Behavior:**
- **As \( x \to \infty \):** \( h(x) \) approaches 1 from above.
- **As \( x \to -\infty \):** \( 2^{-x} = 2^{|x|} \to \infty \), so \( h(x) \to \infty \).
4. **Additional Points:**
- \( x = 1 \):
\[
h(1) = 4 \cdot 2^{-1} + 1 = 4 \cdot \frac{1}{2} + 1 = 2 + 1 = 3 \quad \Rightarrow (1, 3)
\]
- \( x = -1 \):
\[
h(-1) = 4 \cdot 2^{1} + 1 = 4 \cdot 2 + 1 = 8 + 1 = 9 \quad \Rightarrow (-1, 9)
\]
**Graph Description:**
- Start at the \( y \)-intercept \( (0, 5) \).
- The graph decreases as \( x \) increases, approaching the horizontal asymptote \( y = 1 \).
- The graph increases without bound as \( x \) decreases, moving towards infinity.
- Plot additional points like \( (1, 3) \) and \( (-1, 9) \) to guide the shape.
**Visual Representation:**
```
y
|
10| *
9| (-1,9)
|
7|
5|* (0,5)
3| * (1,3)
1|-------------------------> x
0 1 2 3
```
- The line \( y = 1 \) serves as a horizontal asymptote.
- The curve approaches this line as \( x \) increases.
---
### **5.2.4 Describe the transformation from \( h \) to \( g \) if \( g(x) = 4 \cdot \left(2^{-x} + 2\right) \).**
**Solution:**
Let's compare the two functions:
- **Original Function:**
\[
h(x) = 4 \cdot 2^{-x} + 1
\]
- **Transformed Function:**
\[
g(x) = 4 \cdot \left(2^{-x} + 2\right) = 4 \cdot 2^{-x} + 8
\]
**Transformation Steps:**
1. **From \( h(x) \) to \( g(x) \):**
- Start with \( h(x) = 4 \cdot 2^{-x} + 1 \).
- Add 2 inside the parentheses and distribute the 4:
\[
g(x) = 4 \cdot 2^{-x} + 8
\]
- This effectively adds \( 8 - 1 = 7 \) to the original \( h(x) \).
2. **Interpretation:**
- The addition of 7 shifts the entire graph of \( h(x) \) **vertically upwards by 7 units**.
**Answer:**
Function \( g \) is obtained by shifting \( h \) vertically upward by 7 units. In other words, \( g(x) \) is the graph of \( h(x) \) moved up seven units on the \( y \)-axis.
Reviewed and approved by the UpStudy tutoring team
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