I see that you have uploaded an image with several math exercises. I will help you solve these problems one by one. Let's begin by tackling each question in order.

1. Determine the result of the transformation:

$y = 2x + 6 \quad \text{shifted 8 units down.}$ Solution: Shifting the function down by 8 units affects the constant term: $y = 2x + 6 - 8 = 2x - 2.$

2. Determine the result of translating point $H = (2, -3)$ by the vector $T = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$.

Solution: Translation means adding the components of the vector $T$ to the coordinates of point $H$: $H' = (2 + 4, -3 + (-2)) = (6, -5).$

3. The graph of $y = x^2 + bx$ is shifted 2 units to the right. Determine the equation after this transformation.

Solution: A shift of 2 units to the right corresponds to replacing $x$ with $(x - 2)$ in the equation: $y = (x - 2)^2 + b(x - 2).$ Expanding: $y = (x^2 - 4x + 4) + b(x - 2) = x^2 + (b - 4)x + (4 - 2b).$ So, the transformed equation is: $y = x^2 + (b - 4)x + (4 - 2b).$

4. Determine the reflection of point $A(2, 3)$ with respect to various axes and lines:

• a) With respect to the x-axis: Reflection over the x-axis changes the sign of the y-coordinate: $A' = (2, -3).$
• b) With respect to the y-axis: Reflection over the y-axis changes the sign of the x-coordinate: $A' = (-2, 3).$
• c) With respect to the line $y = x$: Swap the coordinates of the point: $A' = (3, 2).$
• d) With respect to the line $y = -x$: Swap and negate the coordinates: $A' = (-3, -2).$
• e) With respect to the origin (0, 0): Negate both coordinates: $A' = (-2, -3).$
• f) With respect to the line $x = 2$: The reflection across $x = 2$ means the x-coordinate is unchanged: $A' = (2, 3).$
• g) With respect to the line $y = 1$: The reflection across $y = 1$ changes the y-coordinate according to the line: $A' = (2, 2 \times 1 - 3) = (2, -1).$

5. Given $f(x) = x^2 + x - 2$, find the reflection of $f(x)$:

• a) Across the x-axis: Reflection over the x-axis changes the sign of the function: $f'(x) = -(x^2 + x - 2) = -x^2 - x + 2.$
• b) Across the y-axis: Reflection over the y-axis means replacing $x$ with $-x$: $f'(x) = (-x)^2 + (-x) - 2 = x^2 - x - 2.$

Do you need detailed explanations for any of these steps or solutions?

Here are five related questions to expand your understanding:

1. How do we perform a vertical stretch or compression on the function $y = 2x + 6$?
2. How would the result change if the translation vector $T$ in question 2 were $T = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$?
3. What effect does shifting the graph of $y = x^2$ by 5 units to the left have on its equation?
4. How would the reflection of $A(2, 3)$ look with respect to the line $y = x + 1$?
5. What happens to a function $f(x)$ if it undergoes a horizontal stretch by a factor of 2?

Tip: When reflecting a point or graph, always identify which coordinate or expression is changing to avoid confusion!