How do you graph ${(x - 3)}^{3} + {(y + 2)}^{2} = 4$ $(x-3)^3 + (y+2)^2 = 4$ on a coordinate graph?

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How do you graph ${(x - 3)}^{3} + {(y + 2)}^{2} = 4$ $(x-3)^3 + (y+2)^2 = 4$ on a coordinate graph?

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Answer 1 · 2015-09-19T14:09:56

Refer to explanation

Explanation:

This is a circle .The general formula is

${(x - x_{o})}_{o}^{2} + {(y - y_{o})}_{o}^{2} = R^{2}$ $(x-x_o)^2+(y-y_o)^2=R^2$ where $(x_{o}, y_{o})$ $(x_o,y_o)$ is the cirlce centre
and R is the radius.

In our case the centre of circle is $(3, - 2)$ $(3,-2)$ and the radius is R=2

Hence the graph is graph{(x-3)^2+(y+2)^2=4 [-8.89, 8.885, -4.444, 4.44]}

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How do you graph ${(x - 3)}^{3} + {(y + 2)}^{2} = 4$ $(x-3)^3 + (y+2)^2 = 4$ on a coordinate graph?

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Explanation:

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How do you graph (x−3)3+(y+2)2=4(x-3)^3 + (y+2)^2 = 4 on a coordinate graph?

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Explanation:

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How do you graph ${(x - 3)}^{3} + {(y + 2)}^{2} = 4$ $(x-3)^3 + (y+2)^2 = 4$ on a coordinate graph?