Let's address the questions:

25(b)

The given function is $q(x) = -f(x-2) + 3$.

This involves a series of transformations on the graph of $f(x)$:

1. Horizontal shift: $x-2$ shifts the graph 2 units to the right.
2. Reflection: The negative sign (-) reflects the graph over the x-axis.
3. Vertical shift: $+3$ shifts the graph 3 units upward.

Steps for sketching:

• Take the key points on the original $f(x)$, apply the transformations sequentially, and plot the modified points.
• Reflect the shape over the x-axis after shifting horizontally.
• Finally, raise the entire graph by 3 units.

Domain and Range:

• The domain of $f(x)$ is $(-\infty, \infty)$, so the domain of $q(x)$ is also $(-\infty, \infty)$.
• If the range of $f(x)$ is $[-5, 6]$, after reflection and vertical shift, the range of $q(x)$ becomes: $\text{Range of } q(x): [-3, 8].$

28(c)

The given function is $p(x) = 2x^2 - 4x + 3$.

Step 1: Rewrite in vertex form To identify transformations, rewrite the quadratic function by completing the square: $p(x) = 2(x^2 - 2x) + 3.$ $p(x) = 2\left((x-1)^2 - 1\right) + 3.$ $p(x) = 2(x-1)^2 - 2 + 3.$ $p(x) = 2(x-1)^2 + 1.$

This is now in the form $a(x-h)^2 + k$, where:

• $a = 2$ (vertical stretch by a factor of 2),
• $h = 1$ (horizontal shift 1 unit to the right),
• $k = 1$ (vertical shift 1 unit upward).

Step 2: Vertex The vertex is $(1, 1)$.

Step 3: X-Intercepts To find the x-intercepts, set $p(x) = 0$: $2x^2 - 4x + 3 = 0.$ Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 2, b = -4, c = 3.$ $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(3)}}{2(2)}.$ $x = \frac{4 \pm \sqrt{16 - 24}}{4}.$ $x = \frac{4 \pm \sqrt{-8}}{4}.$ No real solutions, so there are no x-intercepts.

Step 4: Y-Intercept Set $x = 0$: $p(0) = 2(0)^2 - 4(0) + 3 = 3.$ The y-intercept is $(0, 3)$.

Step 5: Domain and Range

• Domain: All real numbers, $(-\infty, \infty)$.
• Range: Since the vertex is at $(1, 1)$ and the parabola opens upwards, the range is $[1, \infty)$.

Summary of Results:

1. For $q(x) = -f(x-2) + 3$:

   • Domain: $(-\infty, \infty)$,
   • Range: $[-3, 8]$.

2. For $p(x) = 2x^2 - 4x + 3$:

   • Vertex: $(1, 1)$,
   • No x-intercepts,
   • Y-intercept: $(0, 3)$,
   • Domain: $(-\infty, \infty)$,
   • Range: $[1, \infty)$.

Do you want me to sketch these graphs or explain any steps in more detail?

5 Related Questions:

1. How do vertical transformations affect the domain and range of a function?
2. Can a quadratic function have no x-intercepts? Why or why not?
3. How does reflecting a graph over the x-axis affect its range?
4. What is the general method for completing the square in quadratic equations?
5. How can transformations be applied to other types of functions like exponential or logarithmic graphs?

Tip: Always start transformations with horizontal shifts before applying reflections and vertical shifts for accurate graphing!