Question

put the quadratic into vertex form and state the coordinates of the vertex.
$y = x^2 + 14x + 40$
answer attempt 1 out of 2
vertex form: $y = \square$
vertex: $(\square,\square)$

Explanation:

Step1: Complete the square for \(x^2 + 14x\)

To complete the square, take half of the coefficient of \(x\) (which is \(14\)), square it. Half of \(14\) is \(7\), and \(7^2 = 49\). So we rewrite the quadratic as:
\(y = x^2 + 14x + 49 - 49 + 40\)

Step2: Rewrite as a perfect square and simplify

The first three terms \(x^2 + 14x + 49\) form a perfect square \((x + 7)^2\). Then simplify the constants:
\(y=(x + 7)^2 - 49 + 40=(x + 7)^2 - 9\)
The vertex form of a quadratic is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Here, \(h=-7\) (since it's \(x - (-7)\)) and \(k = -9\).

Answer:

Vertex Form: \(y=(x + 7)^2 - 9\)
Vertex: \((-7, -9)\)