Given: $$ h(x)=4\left(2^{-x}\right)+1 $$ 5.2.1 Determine the coordinates of the $$ y $$-intercept of $$ h $$. 5.2.2 Explain why $$ h $$ 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 $$ h $$. 5.2.4 Describe the transformation from $$ h $$ to $$ g $$ if $$ g(x)=4\left(2^{-x}+2\right) $$.

Question

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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.
- 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.

Given:
5.2.1 Determine the coordinates of the -intercept of .
5.2.2 Explain why does not have an -intercept.
5.2.3 Draw a sketch graph of , clearly showing all asymptotes, intercepts with the axes and at least one other point on .
5.2.4 Describe the transformation from to if .

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Given: 5.2.1 Determine the coordinates of the -intercept of . 5.2.2 Explain why does not have an -intercept. 5.2.3 Draw a sketch graph of , clearly showing all asymptotes, intercepts with the axes and at least one other point on . 5.2.4 Describe the transformation from to if .

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Latest Pre Calculus Questions

Given:
5.2.1 Determine the coordinates of the -intercept of .
5.2.2 Explain why does not have an -intercept.
5.2.3 Draw a sketch graph of , clearly showing all asymptotes, intercepts with the axes and at least one other point on .
5.2.4 Describe the transformation from to if .