MedVision ad

curve sketching help (1 Viewer)

Joined
Sep 20, 2015
Messages
43
Gender
Male
HSC
2015
hey guys for this question, i know how to find the horizontal asymptote, i know its symmetrical about the y axis, i know it doesnt go below the x axis the origin and i know as x approaches plus-munis infinity, y approaches 2 but how do you determine the shape?
 

Attachments

kawaiipotato

Well-Known Member
Joined
Apr 28, 2015
Messages
463
Gender
Undisclosed
HSC
2015
That's basically all you need. Except since you know it's reflected about the y-axis, see where it lies ON the y-axis. Sub x = 0, gives you y = 0. So it starts at origin. Unsure how to explain this part, but it has to be a stationary point at the origin, as symmetrical and differentiable curves like this don't have a critical point at the point of symmetry. I may be wrong tho. You'd verify this by calculus, but the question said not to use it so just do it on a spare piece of paper
 
Last edited:
Joined
Sep 20, 2015
Messages
43
Gender
Male
HSC
2015
That's basically all you need. Except since you know it's reflected about the y-axis, see where it lies ON the y-axis. Sub x = 0, gives you y = 0. So it starts at origin. Unsure how to explain this part, but it has to be a stationary point at the origin, as symmetrical and differentiable curves like this don't have a critical point at the point of symmetry. I may be wrong tho. You'd verify this by calculus, but the question said not to use it so just do it on a spare piece of paper
got it, thanks :)
 
Joined
Sep 20, 2015
Messages
43
Gender
Male
HSC
2015
That's basically all you need. Except since you know it's reflected about the y-axis, see where it lies ON the y-axis. Sub x = 0, gives you y = 0. So it starts at origin. Unsure how to explain this part, but it has to be a stationary point at the origin, as symmetrical and differentiable curves like this don't have a critical point at the point of symmetry. I may be wrong tho. You'd verify this by calculus, but the question said not to use it so just do it on a spare piece of paper
got it, thanks.
oh but what do u mean by "critical pt"?
 

kawaiipotato

Well-Known Member
Joined
Apr 28, 2015
Messages
463
Gender
Undisclosed
HSC
2015
got it, thanks.
oh but what do u mean by "critical pt"?
A critical point is when the gradient of the tangent is undefined or "vertical". An example would be at the point (0,0) for y = sqrtx . It initially starts vertical at the origin
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,390
Gender
Male
HSC
2006
A critical point is when the gradient of the tangent is undefined or "vertical". An example would be at the point (0,0) for y = sqrtx . It initially starts vertical at the origin
A critical point also includes points where the derivative is zero.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top