Question

for the quadratic function defined, (a) write the function in the form \\(p(x) = a(x - h)^2 + k\\), (b) give the vertex of the parabola, and (c) graph the function.

\\(p(x) = x^2 - 6x - 4\\)

a. \\(p(x) = \square\\)
(simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Complete the square

To convert \(P(x) = x^2 - 6x - 4\) into vertex form \(P(x) = a(x-h)^2 + k\), we complete the square on the \(x\)-terms.
Take half of the \(x\)-coefficient, \(-6\), which is \(-3\), and square it to get \(9\).
Add and subtract \(9\) inside the expression:

$$ P(x) = (x^2 - 6x + 9) - 9 - 4 $$

Factor the perfect square trinomial:

$$ P(x) = (x-3)^2 - 13 $$

Identify the vertex

The vertex form of a parabola is \(P(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex.
Comparing \(P(x) = (x-3)^2 - 13\) to the standard vertex form:

$$ h = 3, \quad k = -13 $$

Thus, the vertex of the parabola is \((3, -13)\).

Determine graphing points

To graph the function, we find additional points on the parabola.
Let's evaluate \(P(x)\) at \(x = 0\) and \(x = 6\) (symmetric about \(x = 3\)):

$$ P(0) = (0-3)^2 - 13 = 9 - 13 = -4 $$

$$ P(6) = (6-3)^2 - 13 = 9 - 13 = -4 $$

So, the parabola passes through \((3, -13)\), \((0, -4)\), and \((6, -4)\).

Answer:

Question 1

a. \(P(x) =\) <blank>\((x-3)^2-13\)</blank>

Question 2

b. The vertex is <blank>\((3, -13)\)</blank>.

Question 3

c. The graph of the function is a parabola with its vertex at \((3, -13)\) opening upwards, passing through the y-intercept at \((0, -4)\).