Question

1. solve |x + 3| - 1 = (x + 2)² by graphing.

x =
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Explanation:

Step1: Understand the Graphs

We have two functions: \( y = |x + 3| - 1 \) (absolute value function) and \( y=(x + 2)^2 \) (parabola). The solutions to \( |x + 3| - 1=(x + 2)^2 \) are the x - coordinates of the intersection points of the two graphs.

Step2: Analyze the Intersection Points

From the graph, we can see that the two graphs intersect at two points. By looking at the grid and the shape of the graphs (the absolute - value graph is a V - shaped graph and the parabola is a U - shaped graph), we can find the x - values of the intersection points.

We can also solve the equation algebraically to verify:
First, expand the right - hand side: \( (x + 2)^2=x^{2}+4x + 4 \)
The left - hand side is \( |x + 3|-1 \)

Case 1: When \( x+3\geq0 \) (i.e., \( x\geq - 3 \)), the equation becomes \( x + 3-1=x^{2}+4x + 4 \)
Simplify: \( x + 2=x^{2}+4x + 4 \)
Rearrange to standard quadratic form: \( x^{2}+3x + 2 = 0 \)
Factor: \( (x + 1)(x+2)=0 \)
So \( x=-1 \) or \( x=-2 \)

Case 2: When \( x + 3\lt0 \) (i.e., \( x\lt - 3 \)), the equation becomes \( -(x + 3)-1=x^{2}+4x + 4 \)
Simplify: \( -x-3 - 1=x^{2}+4x + 4 \)
\( -x-4=x^{2}+4x + 4 \)
Rearrange to standard quadratic form: \( x^{2}+5x + 8 = 0 \)
The discriminant of this quadratic \( \Delta=b^{2}-4ac=25-32=-7\lt0 \), so no real solutions in this case.

From the graph, we can also visually identify the intersection points. The vertex of the parabola \( y=(x + 2)^2 \) is at \( (-2,0) \), and the vertex of the absolute - value function \( y = |x + 3|-1 \) is at \( (-3,-1) \). By looking at the intersection points on the graph, we find that the x - values of the intersection points are \( x=-2 \) and \( x=-1 \)

Answer:

\( x=-2 \), \( x = - 1 \)