Question

match each quadratic function to its graph.
$f(x) = -2(x + 3)^2 - 1$
$f(x) = -2(x + 3)^2 + 1$
$f(x) = 2(x + 3)^2 + 1$
$f(x) = 2(x - 3)^2 + 1$
(graph with a parabola opening upwards, y-axis labeled with 1 - 7, x-axis not fully labeled, the parabola has a vertex near y=1)

Explanation:

Step1: Analyze the parabola's direction

The graph opens upward, so the coefficient of the squared term should be positive. Among the functions, \( f(x)=2(x + 3)^2+1 \) and \( f(x)=2(x - 3)^2+1 \) have positive coefficients (\( a = 2>0 \)).

Step2: Find the vertex form

The vertex form of a parabola is \( f(x)=a(x - h)^2+k \), where \((h,k)\) is the vertex. For the graph, the vertex seems to be at \( x=-3 \) (since the parabola is shifted left) and \( y = 1 \) (from the graph's minimum point around \( y = 1 \)).

Check the functions:

• \( f(x)=2(x + 3)^2+1 \) has \( h=-3 \), \( k = 1 \) (since \( (x+3)=(x-(-3)) \)).
• \( f(x)=2(x - 3)^2+1 \) has \( h = 3 \), which would shift the parabola to the right, not left as in the graph.

So the function matching the graph (opening upward, vertex at \( (-3,1) \)) is \( f(x)=2(x + 3)^2+1 \).

Answer:

\( f(x)=2(x + 3)^2+1 \)