4x^2+4y^2=64.
what are the domain and range?
my answer is domain;all real numbers
range;-4<=4<=4
this is a circle, centered at the origin
the domain and range are the same
thank you
To determine the domain and range of the equation 4x^2 + 4y^2 = 64, you need to understand the properties of an ellipse equation.
First, let's rearrange the equation to get it into standard form:
Divide the entire equation by 64:
x^2/4 + y^2/16 = 1.
Now, we can identify the key properties of this equation.
For an ellipse in standard form (x^2/a^2 + y^2/b^2 = 1), the major axis is parallel to the x-axis if a > b, and vice versa if b > a.
In this case, the coefficient of x^2 (a^2) is smaller than the coefficient of y^2 (b^2), so the major axis is parallel to the y-axis.
The domain represents the values that x can take. Since the major axis is parallel to the y-axis, there are no restrictions on x. Thus, the domain is indeed all real numbers.
The range represents the values that y can take. In this case, since the major axis is along the y-axis, the range will be limited by the length of the minor axis.
For an ellipse in standard form (x^2/a^2 + y^2/b^2 = 1), the length of the minor axis is equal to 2b. Therefore, the range in this case will be determined by b, which is the square root of the coefficient of y^2. In our equation, b = sqrt(16) = 4.
So, the correct answer is:
Domain: All real numbers
Range: -4 ≤ y ≤ 4