2x^3y^-3*2x^-1y^3 Someone help me with my math.
1 answer:
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X^2 + y^2 - x - 2*y = 0
To find both coordinates and radius we need to make this equation in circle form:
(x-a)^2 + (y-b)^2 = r^2
x^2 - 2*1/2*x + 1/4 - 1/4 + y^2 - 2*1*y + 1 - 1 = 0
Here we are adding and subtracting numbers in order to get square binomial.
(x - 1/2)^2 + (y-1)^2 = 5/4
coordinates of center are (1/2,1) and the radius is √
To find both coordinates and radius we need to make this equation in circle form:
(x-a)^2 + (y-b)^2 = r^2
x^2 - 2*1/2*x + 1/4 - 1/4 + y^2 - 2*1*y + 1 - 1 = 0
Here we are adding and subtracting numbers in order to get square binomial.
(x - 1/2)^2 + (y-1)^2 = 5/4
coordinates of center are (1/2,1) and the radius is √
<h3>Answer: 
</h3>
The -3 is not in the exponent
Explanation:
The parent function is . Plugging in x = 0 leads to y = 1. So the point (0,1) is on the f(x) curve. Going from (0,1) to (0,-2) is a vertical shift of 3 units downward. To represent this shift, we tack on a "-3" at the end of the f(x) function.
You could look at other points as well, but I find working with x = 0 is easiest.
As a check, plugging x = 0 into g(x) leads to...
This confirms our answer.