To rewrite the quadratic function [tex]\( f(x) = 2x^2 + 4x + 4 \)[/tex] in the form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], we need to complete the square. Following a detailed, step-by-step solution:
1. Identify the coefficient [tex]\(a\)[/tex]:
The quadratic term coefficient [tex]\(a\)[/tex] is 2.
2. Complete the square:
- First, factor out the coefficient of [tex]\(x^2\)[/tex] from the quadratic and linear terms:
[tex]\[
f(x) = 2(x^2 + 2x) + 4
\][/tex]
- Next, to complete the square inside the parentheses, we need to add and subtract the square of half the coefficient of [tex]\(x\)[/tex]. The coefficient of [tex]\(x\)[/tex] inside the parentheses is 2, and half of this is 1. Squaring 1 gives us 1. So, we add and subtract 1 inside the parentheses:
[tex]\[
f(x) = 2(x^2 + 2x + 1 - 1) + 4
\][/tex]
- Simplify inside the parentheses by grouping the perfect square trinomial and the constant term:
[tex]\[
f(x) = 2((x + 1)^2 - 1) + 4
\][/tex]
- Distribute the 2 inside the parentheses:
[tex]\[
f(x) = 2(x + 1)^2 - 2 + 4
\][/tex]
- Finally, combine the constant terms:
[tex]\[
f(x) = 2(x + 1)^2 + 2
\][/tex]
Thus, the quadratic function in the form [tex]\( f(x) = a(x-h)^2 + k \)[/tex] is
[tex]\[
f(x) = 2(x + 1)^2 + 2
\][/tex]
3. Identify the vertex:
The vertex form of the quadratic function [tex]\( f(x) = a(x-h)^2 + k \)[/tex] shows that the vertex [tex]\((h, k)\)[/tex] can be read directly from the equation. Here, the equation is
[tex]\[
f(x) = 2(x + 1)^2 + 2
\][/tex]
From this, we see that:
[tex]\[
h = -1 \, \text{and} \, k = 2 - 4 \Rightarrow k = -2
\][/tex]
Therefore, the vertex of the graph is [tex]\((-1, -2)\)[/tex].
### Final Answer:
Quadratic function in vertex form:
[tex]\[
f(x) = 2(x + 1)^2 + 2
\][/tex]
Vertex:
[tex]\[
(-1, -2)
\][/tex]
