Sideways Parabola Question (1 Viewer)

[frenzal_dude]

With curves in the form y²=4ax

What is the minimum and maximum values?

Are the max/min values always in relation to the x-axis, because in this case i guess there are no min/max values, maybe at x = infinity?

Also, are turning points of a graph always either a max/min? because in y²=4ax the turning point is not a min/max.


In the equation y²=2x

y=2x^1/2
y`=1/sqrt(2x)
0=1/sqrt(2x)
multiply both sides by sqrt(2x) and u get 0=1 which doesnt exist, so does that mean there are no max/min points?

 

[Trev]

You're correct, no maximum or minimum points!

 

[damo676767]

i will call the origon a minimum, as that is the smallest y value posible

 

[borkis04]

u idiot dabe

 

[borkis04]

haha jokin man..hey r u studying 4 any other subject except maths and physics? hows english goin 4 ya so far?

 

[frenzal_dude]

haha hey alex, nah man ive just been doing maths like 6 hours a day! i cant find time for anyting else! im gonna study for english 2 days b4 the test coz i hate it so much, im just gonna concentrate on maths, physics, electronics. german and english are no hoper subs for me anyway.

how bout u?

 

[frenzal_dude]

i will call the origon a minimum, as that is the smallest y value posible

but the origin isnt the smallest y value.

Also another question, so the derivative only finds turning points for graphs where the turning point occurs at a gradient of 0, how would u find the turning points of a graph such as y²=4ax where the turning point occurs at gradient = infinity i guess?

 

[rama_v]

but the origin isnt the smallest y value.

Also another question, so the derivative only finds turning points for graphs where the turning point occurs at a gradient of 0, how would u find the turning points of a graph such as y²=4ax where the turning point occurs at gradient = infinity i guess?

y² = 4ax
x = y²/4a
dx/dy = 2y/4a = y/2a
y/2a = 0
y = 0 (since a>0)

I think thats how you would do it. Or I might be completely wrong lol

Try the parabola (y-1)² = 4(x-2)

The 'stationary point' or whatever you would like to call it, where the gradient is infinity should be the vertex i.e. at (2, 1)

y² - 2y + 1 = 4x - 8
4x = y² - 2y + 9
x = (1/4)(y² - 2y + 9)
dx/dy = (1/4)(2y -2)
(1/4)(2y-2) = 0
2y - 2 = 0
2y = 2
y = 1
yeah, that makes sense..hmm

 

[Trev]

but the origin isnt the smallest y value.

Also another question, so the derivative only finds turning points for graphs where the turning point occurs at a gradient of 0, how would u find the turning points of a graph such as y²=4ax where the turning point occurs at gradient = infinity i guess?

There are no turning points!

 

[borkis04]

Man im so lost..this may as well be in swahili for all 9i know :S haha..yeh mate been studying english and geo!..ill ahh..leave maths for the heavens i htink..ill just study a few days b4hand n see how i go..i hav 4 days between english n maths like everrtybody else..plentry of time

haha
good luck to all for maths! i know u guys oprobably dont need it but i certainly do!

haha

ooh yeah cya at the social davo!

 

[frenzal_dude]

will we ever be asked something like this in 2u?

 

[insert-username]

Not sure about the max/min values, but the parabola y^2 = 4ax is certainly part of the 2 unit course.


I_F

 

[frenzal_dude]

so would the origin in that case be aturning point? or is a turning point only ever to do with the x axis?

 

[jake2.0]

it doesn't have a turning point, as a turning point (or inflection (sp?)) occurs when dy/dx=0 right?

y² = 4ax
2y.dy/dx = 4a
dy/dx = 4a/y

which can't equall 0, hence no turning pt.

EDIT: what the hell?? i posted after "word"

 

[word.]

y² = 2x and y = Sqrt{2x} are different.

y² = 2x isn't a function, and y = Sqrt{2x} doesn't have any turning points.

Whether (0,0) is considered a turning point in the relation y² = 2x is beyond the scope of the course.

 

[acmilan]

The equations isnt even a function, i think that eliminates turning points (i think, have to get back on that)