• Want to level up your HSC prep for Maths?
    Click here for details and register for the BoS Trials now!

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
462
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
462
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
8,496
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