Let's solve the problem of converting the given functions into vertex form, which is expressed as:

$P(x) = a(x-h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

Part a) $P(x) = x^2 - 2x - 7$

1. Factor out $a = 1$ (if necessary, but here it remains as 1).
2. Complete the square for $x^2 - 2x$:
   • Take half of the coefficient of $x$ ($-2$) and square it:
     $\left(\frac{-2}{2}\right)^2 = 1$
   • Add and subtract this value inside the equation to complete the square: $P(x) = (x^2 - 2x + 1) - 1 - 7$
   • Simplify: $P(x) = (x - 1)^2 - 8$

Vertex form:
$P(x) = (x - 1)^2 - 8$
Vertex: $(1, -8)$

Part b) $P(x) = -x^2 + 6x + 2$

1. Factor out $a = -1$ from the $x^2$ and $x$ terms: $P(x) = -\left(x^2 - 6x\right) + 2$
2. Complete the square for $x^2 - 6x$:
   • Take half of the coefficient of $x$ ($-6$) and square it:
     $\left(\frac{-6}{2}\right)^2 = 9$
   • Add and subtract this value inside the equation: $P(x) = -\left(x^2 - 6x + 9 - 9\right) + 2$
   • Rewrite: $P(x) = -\left((x - 3)^2 - 9\right) + 2$
   • Distribute the $-1$ and simplify: $P(x) = -(x - 3)^2 + 9 + 2$ $P(x) = -(x - 3)^2 + 11$

Vertex form:
$P(x) = -(x - 3)^2 + 11$
Vertex: $(3, 11)$

Final Answer:

• a) $P(x) = (x - 1)^2 - 8$
• b) $P(x) = -(x - 3)^2 + 11$

Would you like further clarification or a step-by-step explanation of any part?

Related Questions:

1. What is the purpose of completing the square when converting to vertex form?
2. How do you determine the vertex of a quadratic function from its standard form?
3. Can the vertex form be converted back into standard form? If so, how?
4. How does the sign of $a$ in $P(x) = a(x-h)^2 + k$ affect the graph's direction?
5. What are other methods to find the vertex besides completing the square?

Tip:

Always double-check your calculations, especially when factoring or distributing terms, to avoid minor errors.