Question

Transform the quadratic equation from standard form to vertex form and state the vertex point.

y=x2+10x+12

(1 point)
Responses

y=(x+5)2−37
and (−5,−37)
y equals left parenthesis x plus 5 right parenthesis squared minus 37 and left parenthesis negative 5 comma negative 37 right parenthesis

y=(x+5)2−13
and (5,−13)
y equals left parenthesis x plus 5 right parenthesis squared minus 13 and left parenthesis 5 comma negative 13 right parenthesis

y=(x+5)2+13
and (−5,−13)
y equals left parenthesis x plus 5 right parenthesis squared plus 13 and left parenthesis negative 5 comma negative 13 right parenthesis

y=(x+5)2−13
and (−5,−13)

Answer 1

To convert the quadratic equation from standard form \( y = x^2 + 10x + 12 \) to vertex form \( y = a(x-h)^2 + k \), we will complete the square.

1. Start with the standard form:
\[
y = x^2 + 10x + 12
\]

2. Focus on the quadratic and linear terms \( x^2 + 10x \):
- To complete the square, take the coefficient of \( x \) (which is 10), halve it (getting 5), and then square it (getting 25).

3. Rewrite the quadratic part:
\[
y = (x^2 + 10x + 25) - 25 + 12
\]
\[
y = (x + 5)^2 - 13
\]

4. Now, the equation is in vertex form \( y = (x + 5)^2 - 13 \).

The vertex of the parabola is given by the point \((h, k)\), which in this case is:
- \( h = -5 \)
- \( k = -13 \)

Thus, the vertex is \((-5, -13)\).

So the final answer is:
\[
y = (x + 5)^2 - 13 \quad \text{and} \quad (-5, -13)
\]