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for what value (s) of k does the following system of equations have 1 solution? y=(x+2)^(2)-1 and y=k

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Final answer:

For the system of equations to have exactly one solution, the value of k must be where the horizontal line touches the vertex of the parabola. This occurs at k=-1, as the parabola y=(x+2)^2-1 has its vertex at (-2, -1).

Step-by-step explanation:

To determine for what value(s) of k the system of equations y=(x+2)^2-1 and y=k has exactly one solution, we need to understand that the first equation is a parabola that opens upwards, and the second equation is a horizontal line.

The parabola has a vertex at (-2, -1). The coordinate of the vertex suggests that if k is less than -1, the horizontal line y=k would not intersect the parabola, meaning there would be no solutions. If k is exactly equal to -1, the horizontal line would touch the vertex of the parabola, resulting in exactly one solution. For k values greater than -1, the horizontal line would intersect the parabola at two points, resulting in two solutions.

Therefore, the value of k for which there is exactly one solution is k=-1.

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