The range of the function y=x^2 would be y is greater than or equal to 0 in this case. So pretty much just find the vertex of the function and what ever the y coordinate is set that as the lowest number for the range.
y=x^2
помогите плиз сократить дробь 3y^2-8y-6/4-9y^2
The range, usually of a function, is the set of value that the function can take. The integral range is a subset of the range consisting of integer values that the function can take.
4xy + x3y + yx2 + yx + 3yx = x3y + x2y + 8xy = (xy)(x2y + x + 8)
No, but any parabola can be transformed into the form y = x^2.
y=x^2
upward
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y - x = 2 y= -2x + 1
y=x2-6x+9
If you're looking to factor it, you can do it like so: d2ydx2 - yx2 = d3yx2 - yx2 = yx2(d3 - 1)
10
помогите плиз сократить дробь 3y^2-8y-6/4-9y^2
As shown, the function has neither range nor domain.
The range, usually of a function, is the set of value that the function can take. The integral range is a subset of the range consisting of integer values that the function can take.
4xy + x3y + yx2 + yx + 3yx = x3y + x2y + 8xy = (xy)(x2y + x + 8)
range TPate