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We are tasked with analyzing the graphs of the functions \( f(x) = px^2 + qx + r \) (a parabola) and \( g(x) = -2x + 6 \) (a linear function) based on the diagram provided. Here's how we can approach each part:

### Part (a) Find \( B \) (the x-coordinate where the line \( g(x) \) intersects the x-axis).

- \( g(x) = -2x + 6 \) intersects the x-axis when \( g(x) = 0 \). 
  Setting \( -2x + 6 = 0 \):
  \[
  -2x = -6 \quad \Rightarrow \quad x = 3.
  \]
  So, \( B = 3 \).

### Part (b) Find \( C \) (the point of intersection between the parabola \( f(x) \) and the line \( g(x) \)).

- To find \( C \), we need to set \( f(x) = g(x) \).
  \[
  px^2 + qx + r = -2x + 6.
  \]
  This simplifies to:
  \[
  px^2 + (q + 2)x + (r - 6) = 0.
  \]
  Since we are not yet given the values of \( p \), \( q \), and \( r \), we can't solve it directly yet.

### Part (c) Find \( A \) (the x-coordinate where the parabola intersects the x-axis on the left).

- To find \( A \), we set \( f(x) = 0 \), which leads to solving the quadratic equation \( px^2 + qx + r = 0 \).
  The solutions will give us both points where \( f(x) \) crosses the x-axis, but without \( p, q, r \), we need more information to continue.

### Part (d) Find \( p, q, r \) (the coefficients of the quadratic function \( f(x) = px^2 + qx + r \)).

- We know the maximum of the parabola occurs at \( (1, 12) \), meaning \( f(1) = 12 \) and the derivative \( f'(x) = 0 \) at \( x = 1 \).
  
  1. **Using \( f(1) = 12 \)**:
     \[
     p(1)^2 + q(1) + r = 12 \quad \Rightarrow \quad p + q + r = 12. \tag{1}
     \]
  
  2. **The vertex condition \( f'(x) = 0 \) at \( x = 1 \)**:
     The derivative of \( f(x) \) is \( f'(x) = 2px + q \), and at \( x = 1 \):
     \[
     2p(1) + q = 0 \quad \Rightarrow \quad 2p + q = 0. \tag{2}
     \]
  
  3. **Using \( f(x) = 0 \) at \( A \)**: If we can determine \( A \) and substitute it into \( f(A) = 0 \), it will give another equation in \( p, q, r \). From the diagram, \( A \) appears to be near \( x = -2 \).

We can solve equations (1) and (2) once we gather more information.

### Part (e) Find \( D \) (the y-coordinate where the parabola intersects the y-axis).

- The parabola intersects the y-axis where \( x = 0 \). Substituting \( x = 0 \) into \( f(x) \):
  \[
  f(0) = r.
  \]
  So, \( D = r \).

### Part (f) Find \( P \) (the point where the line intersects the y-axis).

- The line \( g(x) = -2x + 6 \) intersects the y-axis when \( x = 0 \):
  \[
  g(0) = 6.
  \]
  So, \( P = 6 \).

Would you like to delve deeper into solving for \( p, q, r \) or any of the other parts? Let me know if you'd like further details on these approaches!

### Relative Questions:
1. How does the derivative help us find the maximum point of a quadratic function?
2. What is the significance of solving the system of equations for \( p, q, r \)?
3. How would you solve a quadratic equation if given specific values for \( p, q, r \)?
4. Can the symmetry of the parabola provide any clues about the values of \( p, q, r \)?
5. What other information can we extract from the intersection points of the parabola and line?

### Tip:
For quadratic functions, the vertex form \( f(x) = a(x - h)^2 + k \) is a quick way to find the vertex and maximum/minimum values.

The graphs of f(x) = px^2 + qx + r and g(x) = -2x + 6 are sketched. (1, 12) are the coordinates of the maximum value of f. g crosses the x-axis at B. Find: (a) B, (b) C, (c) A, (d) p, q, r, (e) D, (f) P.

Explore how to solve the intersection between a quadratic function f(x) = px^2 + qx + r and a linear function g(x) = -2x + 6, including finding points A, B, C, D, and P. This problem incorporates quadratic roots, derivatives, and maximum value conditions.

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