QUESTION 2 2. $$ \begin{array}{ll} ext { Given: } h(x)=4\left(2^{-x} ight)+1 \ .2 .1 & ext { Determine the coordinates of the } y ext {-intercept of } h \ .2 .2 & ext { Explain why } h ext { does not have an } x ext {-intercept. } \ .2 .3 & \begin{array}{l} ext { Draw a sketch graph of } h ext {, clearly showing all asymptotes, intercepts } \ ext { with the axes and at least one other point on } h\end{array} \ .2 .4 & ext { Describe the transformation from } h ext { to } g ext { if } g(x)=4\left(2^{-s}+2 ight) .\end{array} $$

Question

QUESTION 2 2. $$ \begin{array}{ll}	ext { Given: } h(x)=4\left(2^{-x}ight)+1 \ .2 .1 & 	ext { Determine the coordinates of the } y 	ext {-intercept of } h \ .2 .2 & 	ext { Explain why } h 	ext { does not have an } x 	ext {-intercept. } \ .2 .3 & \begin{array}{l}	ext { Draw a sketch graph of } h 	ext {, clearly showing all asymptotes, intercepts } \ 	ext { with the axes and at least one other point on } h\end{array} \ .2 .4 & 	ext { Describe the transformation from } h 	ext { to } g 	ext { if } g(x)=4\left(2^{-s}+2ight) .\end{array} $$

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Let's solve the problem step by step.

### Part 2.1: Determine the coordinates of the $$ y $$-intercept of $$ h $$

To find the $$ y $$-intercept of the function $$ h(x) = 4(2^{-x}) + 1 $$, we need to evaluate $$ h(0) $$:

$$
h(0) = 4(2^{-0}) + 1 = 4(1) + 1 = 4 + 1 = 5
$$

Thus, the coordinates of the $$ y $$-intercept are $$ (0, 5) $$.

### Part 2.2: Explain why $$ h $$ does not have an $$ x $$-intercept

An $$ x $$-intercept occurs where $$ h(x) = 0 $$. Let's set the equation to zero and solve for $$ x $$:

$$
4(2^{-x}) + 1 = 0
$$

Rearranging gives:

$$
4(2^{-x}) = -1
$$

Since $$ 2^{-x} $$ is always positive for all real $$ x $$, the left side $$ 4(2^{-x}) $$ is also positive. Therefore, it can never equal $$-1$$. This means that $$ h(x) $$ does not cross the $$ x $$-axis, and thus, $$ h $$ does not have an $$ x $$-intercept.

### Part 2.3: Draw a sketch graph of $$ h $$

To sketch the graph of $$ h(x) = 4(2^{-x}) + 1 $$, we need to identify the asymptotes, intercepts, and at least one other point.

1. **Asymptote**: As $$ x \to \infty $$, $$ 2^{-x} \to 0 $$, so $$ h(x) \to 1 $$. Therefore, there is a horizontal asymptote at $$ y = 1 $$.
2. **Intercepts**: We already found the $$ y $$-intercept at $$ (0, 5) $$.
3. **Another point**: Let's find $$ h(1) $$:

$$
h(1) = 4(2^{-1}) + 1 = 4(0.5) + 1 = 2 + 1 = 3
$$

So, another point is $$ (1, 3) $$.

Now, we can sketch the graph. The graph starts from the $$ y $$-intercept at $$ (0, 5) $$, decreases towards the horizontal asymptote at $$ y = 1 $$, and passes through the point $$ (1, 3) $$.

### Part 2.4: Describe the transformation from $$ h $$ to $$ g $$

The function $$ g(x) = 4(2^{-s} + 2) $$ can be rewritten as:

$$
g(x) = 4(2^{-x}) + 8
$$

This shows that $$ g(x) $$ is derived from $$ h(x) $$ by shifting the entire graph of $$ h(x) $$ vertically upwards by 7 units (since $$ 8 - 1 = 7 $$).

### Summary of Results

- **Coordinates of the $$ y $$-intercept**: $$ (0, 5) $$
- **No $$ x $$-intercept**: $$ h(x) $$ does not cross the $$ x $$-axis.
- **Graph**: Sketch includes the $$ y $$-intercept at $$ (0, 5) $$, horizontal asymptote at $$ y = 1 $$, and point $$ (1, 3) $$.
- **Transformation**: $$ g(x) $$ is a vertical shift of $$ h(x) $$ upwards by 7 units.
- **$$ y $$-intercept**: $$ (0, 5) $$ - **No $$ x $$-intercept**: The function $$ h(x) $$ does not cross the $$ x $$-axis. - **Graph**: The graph of $$ h(x) $$ has a $$ y $$-intercept at $$ (0, 5) $$, a horizontal asymptote at $$ y = 1 $$, and passes through the point $$ (1, 3) $$. - **Transformation**: $$ g(x) $$ is obtained by shifting $$ h(x) $$ upwards by 7 units.

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