Question

analyzing translated graphs
determine the vertex form of ( g(x) = x^2 + 2x - 1 ). which graph represents ( g(x) )?
four graphs of parabolas on coordinate grids with options as circles below each graph

Explanation:

Step1: Convert to vertex form

To find the vertex form \( g(x) = a(x - h)^2 + k \), complete the square for \( g(x)=x^2 + 2x - 1 \).
Take the coefficient of \( x \), which is \( 2 \), divide by \( 2 \) to get \( 1 \), and square it to get \( 1^2 = 1 \).
Add and subtract this value inside the equation:
\( g(x)=x^2 + 2x + 1 - 1 - 1 \)
The first three terms form a perfect square trinomial: \( (x + 1)^2 \)
So, \( g(x)=(x + 1)^2 - 2 \)
The vertex of the parabola is at \( (h, k)=(-1, -2) \) (since \( h=-1 \) from \( x - h = x + 1 \) and \( k = -2 \)).

Step2: Analyze the graphs

• The first graph: Vertex is not at \( (-1, -2) \) (its vertex seems higher).
• The second graph: Let's check the vertex. The vertex form is \( (x + 1)^2 - 2 \), so when \( x=-1 \), \( y=-2 \). Let's see the second graph: when \( x=-1 \), what's \( y \)? Wait, no, let's check the vertex coordinates. The vertex is \( (-1, -2) \). Let's check the second graph: the parabola opens upwards, vertex at \( (1, -2) \)? No, wait, let's re - check. Wait, the second graph: when \( x = 0 \), \( y=-1 \)? Wait, no, let's use the vertex. The vertex of \( g(x)=(x + 1)^2 - 2 \) is \( (-1, -2) \). Let's look at the fourth graph: the vertex is at \( (-1, -2) \). Wait, let's check each graph:

First graph: Vertex around \( (-1, 2) \) (since it's above the x - axis, y - value positive). Not matching.

Second graph: Vertex around \( (1, -2) \). Not matching.

Third graph: Vertex around \( (1, 2) \). Not matching.

Fourth graph: Vertex at \( (-1, -2) \), which matches the vertex of \( g(x)=(x + 1)^2 - 2 \). Also, the parabola opens upwards (since the coefficient of \( x^2 \) is positive), which all graphs do. So the fourth graph (the last one) represents \( g(x) \).

Answer:

The graph on the right (the fourth graph) represents \( g(x)=x^2 + 2x - 1 \) (after converting to vertex form \( g(x)=(x + 1)^2 - 2 \) with vertex at \( (-1, -2) \)).