Math, asked by nathanj1103, 10 months ago

Which is one of the transformations applied to the graph of f(x)=x2 to change it into the graph of g(x) = –x2 + 16x – 44?

A). The graph of f(x) = x2 is widened.
B). The graph of f(x) = x2 is shifted left 8 units.
C). The graph of f(x) = x2 is shifted down 44 units.
D). The graph of f(x) = x2 is reflected over the x-axis.

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Answers

Answered by Afreenakbar
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D) The graph of f(x) = x^2 is reflected over the x-axis: This option is correct. Reflecting the graph of f(x) = x^2 over the x-axis would change the sign of the y-coordinate of all the points on the graph, making the parabola open downwards, which is the case with the graph of g(x) = –x^2 + 16x – 44.

A) The graph of f(x) = x^2 is widened: This option is incorrect. Widening a graph would result in the increase of the distance between the vertex and the x-axis or the y-axis, which would change the shape of the parabola.

B) The graph of f(x) = x^2 is shifted left 8 units: This option is incorrect. Shifting the graph of f(x) = x^2 left 8 units would mean that the vertex of the parabola would be located at the point (-8, k), where k is the y-coordinate of the vertex. But the equation g(x) = –x^2 + 16x – 44 doesn't have the vertex at that point.

C) The graph of f(x) = x^2 is shifted down 44 units: This option is incorrect. Shifting the graph of f(x) = x^2 down 44 units would mean that the vertex of the parabola would be located at the point (k, -44), where k is the x-coordinate of the vertex. But the equation g(x) = –x^2 + 16x – 44 doesn't have the vertex at that point.

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