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

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.