The domain of y = 2x is [0, +infinity].
(-infinity, infinity)
it is just one tangent interval except sideways from left to right with domain or (-inf,+inf) and the range of (-pi/2,pi/2)
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
-2
Yes, that is a shifted tanX graph, just as you would shift any graft.
(-infinity, infinity)
it is just one tangent interval except sideways from left to right with domain or (-inf,+inf) and the range of (-pi/2,pi/2)
Given the function g(f(x)) = 2-x, you can find the domain as you would with any other function (i.e. it doesn't matter if it's composite). The output, however, has to be a real number. With this function, the domain is all real numbers. If you graph it, you see that the function is defined across the entire graph, wherever you choose to plot it.
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
-2
Any graph with the slope of -1/2
y equals x-4 plus 2 is the same as y = x-2. You just translate the graph of y=x, 2 units to the right, OR 2 down.
Find the domain of the relation then draw the graph.
3
Yes, that is a shifted tanX graph, just as you would shift any graft.
The domain would be (...-2,-1,0,1,2...); the range: (12)
(x^2)^(1/2) equals x, therefore, y = x+4, which has a range and domain of all real numbers. The graph is a straight line, slope of 1, y-intercept of 4. Are you actually saying y = (x^2+4)^(1/2). If so, the range and domain will also be all real numbers because x^2+4 will never result in a negative number.