To solve this problem of determining the equation of \( h(x) \) in vertex form given the table of values, we'll go through a systematic approach.
First, let's recall the forms of quadratic functions:
- Standard form: \( h(x) = ax^2 + bx + c \)
- Vertex form: \( h(x) = a(x - h_v)^2 + k \), where \((h_v, k)\) is the vertex of the parabola.
Given the data points in the table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & h(x) \\
\hline
-3 & -2 \\
\hline
-2 & -3 \\
\hline
-1 & -2 \\
\hline
0 & 1 \\
\hline
1 & 6 \\
\hline
2 & 13 \\
\hline
3 & 22 \\
\hline
\end{array}
\][/tex]
1. Fit a Quadratic Function:
We fit a quadratic function \( h(x) = ax^2 + bx + c \) using the given points.
2. Determine the Coefficients:
The quadratic function that fits these points can be determined, and we find the coefficients to be \( a = 1 \), \( b = 4 \), and \( c = 1 \). Thus, the quadratic function in standard form is:
[tex]\[
h(x) = x^2 + 4x + 1
\][/tex]
3. Convert to Vertex Form:
To convert to vertex form, we use the relationship \( x = -\frac{b}{2a} \) to find the x-coordinate of the vertex. Here \( a = 1 \) and \( b = 4 \):
[tex]\[
h_v = -\frac{4}{2 \cdot 1} = -2
\][/tex]
We then find the corresponding \( y \)-value (k) by substituting \( h_v = -2 \) back into the quadratic equation:
[tex]\[
k = h(-2) = (1)(-2)^2 + 4(-2) + 1 = 4 - 8 + 1 = -3
\][/tex]
Therefore, the vertex form of the quadratic equation is:
[tex]\[
h(x) = 1(x - (-2))^2 + (-3)
\][/tex]
Simplifying, we get:
[tex]\[
h(x) = (x + 2)^2 - 3
\][/tex]
4. Match with Given Options:
We compare our derived vertex form with the given options:
- \( (x+2)^2 - 3 \)
- \( (x+1)^2 - 2 \)
- \( (x-1)^2 + 2 \)
- \( (x-2)^2 + 3 \)
We see that the vertex form \( h(x) = (x + 2)^2 - 3 \) corresponds perfectly with the first option.
Thus, the correct equation in vertex form is:
[tex]\[ h(x) = (x + 2)^2 - 3 \][/tex]
Therefore, the correct choice is:
[tex]\[ \boxed{h(x) = (x + 2)^2 - 3} \][/tex]
