Question

identify the graph of $g(x) = (x + 1)^2 - 1$?
compare the graph to the graph of $f(x) = x^2$.

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For \( g(x)=(x + 1)^2-7 \), we can rewrite \( x + 1 \) as \( x-(-1) \), so \( h=-1 \) and \( k = -7 \). Thus, the vertex of \( g(x) \) is \((-1, -7)\).

Step2: Analyze the parent function \( f(x)=x^2 \)

The parent function \( f(x)=x^2 \) has a vertex at \((0, 0)\), opens upward (since \( a = 1>0 \)).

Step3: Determine transformation

For \( g(x)=(x + 1)^2-7 \), compared to \( f(x)=x^2 \):

• The \( x + 1 \) inside the square means a horizontal shift. Since it's \( x-(-1) \), it's a shift 1 unit to the left.
• The \(-7\) outside the square means a vertical shift 7 units down.

So the graph of \( g(x) \) should have its vertex at \((-1, -7)\), open upward (same as \( f(x) \) since \( a = 1>0 \)). Now, looking at the options, we need to find the parabola with vertex at \((-1, -7)\) and opening upward.

Answer:

The graph of \( g(x)=(x + 1)^2-7 \) has its vertex at \((-1, -7)\), opens upward, and is a transformation of \( f(x)=x^2 \) (shifted 1 unit left and 7 units down). To identify the correct graph, look for the parabola with vertex at \((-1, -7)\) (left 1, down 7 from the origin) and opening upwards.

For the comparison: The graph of \( g(x) \) is the graph of \( f(x)=x^2 \) shifted 1 unit to the left and 7 units down.