Someone please tell me the domain and range. Thank you.
1 answer:
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X² + 8x +y² - 2y -64 =0
We see that the equation has x² and y² . Also we see that coefficients in front of x² and y² are equal. So this is an equation of the circle.
(x² + 8x) +(y² - 2y) -64 =0
(x² + 8x) +(y² - 2y) = 64
We need to complete square for x and y groups, that means it should be written in form (a+b)² or (a-b)².
Expressions in parenthesis we will write as a²+/-2ab+b², to write it after as (a+/-b)², because a²+/-2ab+b² = (a+/-b)²
(x² + 2*4x) +(y² - 2*1y) = 64
(x² + 2*4x+4²) +(y² - 2*1y+1²) = 64+4²+1²
(x+4)² + (y-1)²= 81 Sometimes this is called a standard form of the circle.
(x+4)² + (y-1)²= 9² Sometimes it is required to write like this.
And if you are studying circles,ellipses and hyperbolas, the standard form should look like
We see that the equation has x² and y² . Also we see that coefficients in front of x² and y² are equal. So this is an equation of the circle.
(x² + 8x) +(y² - 2y) -64 =0
(x² + 8x) +(y² - 2y) = 64
We need to complete square for x and y groups, that means it should be written in form (a+b)² or (a-b)².
Expressions in parenthesis we will write as a²+/-2ab+b², to write it after as (a+/-b)², because a²+/-2ab+b² = (a+/-b)²
(x² + 2*4x) +(y² - 2*1y) = 64
(x² + 2*4x+4²) +(y² - 2*1y+1²) = 64+4²+1²
(x+4)² + (y-1)²= 81 Sometimes this is called a standard form of the circle.
(x+4)² + (y-1)²= 9² Sometimes it is required to write like this.
And if you are studying circles,ellipses and hyperbolas, the standard form should look like