Graphing (1 Viewer)

Lukybear · Dec 21, 2009

How to graph this from scratch?

y=x(3+sqrtx)

Do we have to know the general form of it?

Cazic · Dec 21, 2009

You know what the graph for sqrt(x) looks like, right? Well, this thing will be the same basic shape, just increasing quicker.

To get your bearings for this one as opposed to sqrt(x), this thing still starts at (0,0), but passes through (1, 4) and (4, 20) instead of (1,1) and (4, 2). Same shape but.

lychnobity · Dec 21, 2009

Lukybear said:
How to graph this from scratch?

y=x(3+sqrtx)

Do we have to know the general form of it?

No, just know how to do it. ie process

Use multiplication

1) Draw $3 + \sqrt{x}$
2) Draw x
3) Multiply the ordinates

Tofuu · Dec 21, 2009

lychnobity said:
no, just know how to do it. Ie process

use multiplication

1) draw $3 + \sqrt{x}$
2) draw x
3) multiply the ordinates

+1

Cazic · Dec 21, 2009

Better to just get a feel for how multiplying by 'x' changes a graph imo. Doing it pointwise like that without a feeling might be time consuming for no reason.

The Nomad · Dec 22, 2009

You can draw 3 + x^0.5, then to get x(3 + x^0.5): add an x-intercept at x = 0.

khorne · Dec 22, 2009

untouchablecuz said:
$\\y=3x+x^{\frac{3}{2}} \text{ and }\frac{dy}{dx}=3+1.5x^{0.5}\\x\text { must be greater than or equal to zero since } y \text{ is complex for }x<0\\\frac{dy}{dx} \text{ is undefined when }x=0 \text{ meaning that at this point there is a vertical tangent}\\\text{It can be intuitively seen that for this vertical tangent to occur, the curve must be concave down}\\}\text{As }x\to \infty ^+\text{, }y\rightarrow \infty ^+$

use this information to sketch the graph and imo, multiplication/addition/etc of ordinates should be avoided at all costs. too inaccurate/tiresome

Use this method.

SeCKSiiMiNh · Dec 22, 2009

If all else fails, you could always just plot it the good old fashion way

Cazic · Dec 22, 2009

khorne said:
Use this method.

I don't see why.

The derivative is undefined at x = 0 because the function is undefined for x < 0, but that doesn't mean there is going to be a vertical tangent there. In fact, given the derivative he listed, my guess is the function has a tangent line whose slope is very close to 3 when you're very close to 0.

Analysing the function a bit more, a function is concave down on some region if the derivative is decreasing (not the same as negative) on that region. But the derivative is clearly increasing on any region the function is defined on (the square root function in the first derivative is an increasing function "by memory", or consider that the second derivative is always positive). So it's actually concave up.

untouchablecuz · Dec 22, 2009

Cazic said:
I don't see why.

The derivative is undefined at x = 0 because the function is undefined for x < 0, but that doesn't mean there is going to be a vertical tangent there. In fact, given the derivative he listed, my guess is the function has a tangent line whose slope is very close to 3 when you're very close to 0.

Analysing the function a bit more, a function is concave down on some region if the derivative is decreasing (not the same as negative) on that region. But the derivative is clearly increasing on any region the function is defined on (the square root function in the first derivative is an increasing function "by memory", or consider that the second derivative is always positive). So it's actually concave up.

you are correct

i quickly assumed a vertical tangent without taking limits

thanks

Lukybear · Dec 22, 2009

Cazic said:
I don't see why.

The derivative is undefined at x = 0 because the function is undefined for x < 0, but that doesn't mean there is going to be a vertical tangent there. In fact, given the derivative he listed, my guess is the function has a tangent line whose slope is very close to 3 when you're very close to 0.

Analysing the function a bit more, a function is concave down on some region if the derivative is decreasing (not the same as negative) on that region. But the derivative is clearly increasing on any region the function is defined on (the square root function in the first derivative is an increasing function "by memory", or consider that the second derivative is always positive). So it's actually concave up.

One question that i have is why is the vertical tangent always stated, even it isnt really relevant to question.

Also how would we formalise this solution?

untouchablecuz · Dec 23, 2009

$\\y=3x+x^{\frac{3}{2}} \text{, }\frac{dy}{dx}=3+1.5x^{0.5}\\x\geq0 \text { since } y \text{ is complex for }x<0\\\frac{dy}{dx} \text{ is undefined when }x=0 \\\ \text {Taking the limit, }\lim_{x\to 0}(\frac{dy}{dx})=\lim_{x\to 0}(3+1.5x^{0.5})=3\\\therefore y=3x \text{ is tangent to the curve as zero is approached}\\\text{One can see that this tangent conditions that the curve is concave up for all non negative }x \text{ i.e. }\frac{d^2y}{dy^2}>0 \text{ for }x\geq 0\\\text{As }x\to \infty ^+\text{, }y\rightarrow \infty ^+$

ok this is def. correct

GUSSSSSSSSSSSSS · Dec 23, 2009

why shud multiplication of ordinates be avoided at all costs untouchablecuz???
i found its very quick to get the shape of the curve and then with a little testing you can find out the exact points that are necessary

..

khorne · Dec 23, 2009

Cazic said:
I don't see why.

The method outlined is correct, although the working for the question isn't.

Graphing (1 Viewer)

Lukybear

Active Member

Cazic

Member

lychnobity

Active Member

Tofuu

Member

Cazic

Member

The Nomad

Member

khorne

Guest

SeCKSiiMiNh

i'm a fireball in bed

Cazic

Member

untouchablecuz

Active Member

Lukybear

Active Member

untouchablecuz

Active Member

GUSSSSSSSSSSSSS

Active Member

khorne

Guest

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