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PLEASE HELP ME ANSWER THIS QUESTION.

PLEASE HELP ME ANSWER THIS QUESTION.-example-1

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answer:

To write the equation in standard form, we need to complete the square for both the x and y terms. Let's start with the x terms:

-x^2 - 2x + y^2 + 6y - 73 = 0

To complete the square for the x terms, we need to add (2/2)^2 = 1 to both sides of the equation:

-x^2 - 2x + 1 + y^2 + 6y - 73 + 1 = 0 + 1

Simplifying, we have:

-(x^2 + 2x + 1) + y^2 + 6y - 72 = 1

Now, let's do the same for the y terms:

-(x^2 + 2x + 1) + (y^2 + 6y + 9) - 72 = 1 + 9

Simplifying further:

-(x^2 + 2x + 1) + (y^2 + 6y + 9) = 82

Factoring the x and y terms:

-(x + 1)^2 + (y + 3)^2 = 82

This is the standard form of the conic.

To plot the graph, you can use a graphing calculator like Desmos. Set the equation to -(x + 1)^2 + (y + 3)^2 = 82 and plot it.

The vertices of the graph can be found by changing the signs in the equation:

(x + 1)^2 - (y + 3)^2 = 82

The x and y coordinates of the vertices can be found by setting the expressions inside the parentheses to zero:

x + 1 = 0 => x = -1

y + 3 = 0 => y = -3

So, the vertices of the graph are (-1, -3).

real

answered
User Adrien Givry
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