the vertex of f(x) is (2,9)
The vertex of g(x) is to the right 2 units and 5 units down, so its vertex must be (4,4)
since the "shape" has to be the same, we need the same 3 in front
so g(x) = 3(x-4)^2 + 4
domain: the set of real numbers
range: y ≥ 4 , since the parabola opens up
http://www.wolframalpha.com/input/?i=plot+y+%3D+3(x-2)%5E2%2B9+,+y+%3D+3(x-4)%5E2+%2B+4
c) how about h(x) = -3(x-4)^2 + 4
http://www.wolframalpha.com/input/?i=plot+y+%3D+3(x-4)%5E2+%2B+4,+y+%3D+-3(x-4)%5E2+%2B+4
The graph of the function g(x) has the same shape and direction of opening as the graph of f(x)= 3(x-2)^2+9. The graph of g(x) has a vertex that is 2 units to the right and 5 units down from the vertex of the graph of f(x).
A.) Determine an equation of the function g(x)
B.) State the domain and range
C.) Write another function h(x) with the same vertex and shape but whose graph opens in the opposite direction.
If you could show me step by step how to to do, that would be awesome. Thanks
1 answer