Question

sanjay begins to correctly graph the function $f(x) = (x + 1)^2 - 3$. based on the axis of symmetry and the vertex, which graph could be sanjays? two graphs with coordinate planes and marked points/axes are shown, along with radio buttons for selection

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex and the axis of symmetry is \( x = h \).

Step2: Identify vertex and axis of symmetry for \( f(x)=(x + 1)^2 - 3 \)

For the function \( f(x)=(x + 1)^2 - 3 \), we can rewrite it as \( f(x)=1(x - (-1))^2 + (-3) \). So, the vertex \((h, k)\) is \((-1, -3)\) and the axis of symmetry is \( x=-1 \).

Step3: Analyze the given graphs

We need to check which graph has vertex \((-1, -3)\) and axis of symmetry \( x = -1 \). (Note: From the provided options, we assume one of them has vertex \((-1, -3)\) and axis \( x=-1 \), while the others have incorrect vertices/axes. Since the first graph shown has vertex \((1, -3)\) (incorrect \( h \)) and the second has \((-3, 1)\) (incorrect), but likely there is a correct graph with vertex \((-1, -3)\) and axis \( x=-1 \) among the options, which we would identify based on the vertex form analysis.)

Answer:

The graph with vertex \((-1, -3)\) and axis of symmetry \( x = -1 \) (assuming such a graph is among the options, with the correct vertex and axis as derived from the function \( f(x)=(x + 1)^2 - 3 \)).