y=ax^3 + d (1 Viewer)

[xXmuffin0manXx]

the grapih that is set out in y = ax^3 + d

How do you find the equation for this for this graph?

Say, y-intercept is 10 and a point is (2,20)

?

 

[12o9]

sub in both points (o,10) and (2,20) then solve simultaneously =)

 

[xXmuffin0manXx]

ahh right got it..thanks

 

[kaz1]

Alternative method:

d=10 since the y intercept is the constant. Then sub in (2,20) to find a.

 

[xXmuffin0manXx]

isnt that just the second step of what the other guy said ?

thanks anyway

 

[Continuum]

Nope, because the first method relied on simultaneous equations. The one that kaz1 proposed is abit quicker because it gives a value to the constant straight off and just requires you to sub in one point.

With the first method, you can use it whenever. With the second method, you need a point whose x value is 0.

 

[nerdsforever]

Parabolas = boring = useless to learn

 

[xXmuffin0manXx]

Parabolas = boring = useless to learn

it isnt that bad..haha (here i am defending maths)..

but then again..i was away from school for the better part of 2 weeks coz
of an injury which made me to teach the content to myself using the textbook which i think added to the 'fun' if you wanna put it that way..

got a tutor to just make sure i was 100% competent with this today which wll hopefully allow me to catch up..

thanks for all yor responses

 

[Aplus]

Parabolas = boring = useless to learn

Parabolas are good

 

[seano77]

Parabolas = boring = useless to learn

If you don't learn them you are in bad news for year 12 maths.

 

[tommykins]

回复: Re: y=ax^3 + d

Parabolas = boring = useless to learn

HAHAH fuck wait until you hit year 11/12 when parametrics comes in.

 

[Slidey]

Parabolas = boring = useless to learn

Oh, really? Have fun failing HSC and university maths and physics.

And lol, y=ax^3+d isn't a parabola anyway - it's a cubic.