Let's analyze the given expression and rewrite it in the form [tex]\( (x + c)^2 + d \)[/tex].
### Part (a) Completing the square
To complete the square for the expression [tex]\( x^2 + 6x + 11 \)[/tex]:
1. Start with the quadratic expression:
[tex]\[
x^2 + 6x + 11
\][/tex]
2. Take the coefficient of [tex]\( x \)[/tex], which is 6, divide it by 2 and square it:
[tex]\[
\left(\frac{6}{2}\right)^2 = 3^2 = 9
\][/tex]
3. Add and subtract this squared value inside the expression:
[tex]\[
x^2 + 6x + 9 + 11 - 9
\][/tex]
4. Now, rewrite the quadratic and the perfect square trinomial:
[tex]\[
(x^2 + 6x + 9) + 2 = (x + 3)^2 + 2
\][/tex]
So, the expression [tex]\( x^2 + 6x + 11 \)[/tex] can be written as [tex]\( (x + 3)^2 + 2 \)[/tex].
Thus, we have:
[tex]\[
c = 3 \quad \text{and} \quad d = 2
\][/tex]
### Part (b) Finding the coordinates of the turning point
The turning point (or vertex) of the quadratic function [tex]\( y = x^2 + 6x + 11 \)[/tex] is the minimum point of the parabola described by the expression [tex]\( (x + 3)^2 + 2 \)[/tex].
The expression in the form [tex]\( (x + 3)^2 + 2 \)[/tex] reveals that the quadratic achieves its minimum value when the squared term [tex]\( (x + 3)^2 \)[/tex] is zero. This happens when [tex]\( x = -3 \)[/tex].
- At [tex]\( x = -3 \)[/tex]:
[tex]\[
y = (x + 3)^2 + 2 = (0)^2 + 2 = 2
\][/tex]
Therefore, the turning point of the curve [tex]\( y = x^2 + 6x + 11 \)[/tex] is:
[tex]\[
(-3, 2)
\][/tex]
### Summary
(a) The expression [tex]\( x^2 + 6x + 11 \)[/tex] can be written as [tex]\( (x + 3)^2 + 2 \)[/tex] with [tex]\( c = 3 \)[/tex] and [tex]\( d = 2 \)[/tex].
(b) The coordinates of the turning point of the curve [tex]\( y = x^2 + 6x + 11 \)[/tex] are [tex]\( (-3, 2) \)[/tex].
