Answer :
y= 2x^2-3
x= 2y^2-3
x+3=2y^2
(x+3)/2= y^2
y= squareroot [(x+3)/2)] --> inverse equation
Domain: (-3, infinity)
Range: (0, infinity)
y= 3x-1
Let x=0
y=3(0)-1
y=-1, Therefore when x=0,y=-1, so point of graph will be (0.-1)
Let x=1
y= 3(1)-1
y= 2, therefore when x=1, y=2 ...point on graph will be (1,2)
Let x be-1
y= 3(-1-)-1
y=-4, therefore when x=-1, y=-4...points on graph will be (-1,-4)
Hope this helps!!!
x= 2y^2-3
x+3=2y^2
(x+3)/2= y^2
y= squareroot [(x+3)/2)] --> inverse equation
Domain: (-3, infinity)
Range: (0, infinity)
y= 3x-1
Let x=0
y=3(0)-1
y=-1, Therefore when x=0,y=-1, so point of graph will be (0.-1)
Let x=1
y= 3(1)-1
y= 2, therefore when x=1, y=2 ...point on graph will be (1,2)
Let x be-1
y= 3(-1-)-1
y=-4, therefore when x=-1, y=-4...points on graph will be (-1,-4)
Hope this helps!!!
Answer:
The answer would be C) (-1/2,-5/2),(2,5) I believe. :D
