calculus (1 Viewer)

[Mu5hi]

Draw neat sketch of the following graphs
2x^2-4x-6

dx/dy=4x-4

4(x-1) <------------is this right

when i use box for maximum/minimum

do i put x value as 4 or 1? (for the first point)

 

[Timothy.Siu]

stationary point is when y'=0 , so x=1

 

[Mu5hi]

stationary point is when y'=0 , so x=1

only 1 stationary point?

 

[Timothy.Siu]

only 1 stationary point?

yes, quadratics can only have 1 stationary point.
cubics can have 2

 

[Gibbatron]

If you draw a quick graph, you can see its a parabola, and therefore only has one stationary point.

 

[Mu5hi]

yes, quadratics can only have 1 stationary point.
cubics can have 2

If you draw a quick graph, you can see its a parabola, and therefore only has one stationary point.

silly me for missing that, i was just doing all these cubics, and a quadratic came inbetween them,

thnx for the help.

 

[zazzy1234]

Draw neat sketch of the following graphs
2x^2-4x-6

dx/dy=4x-4

4(x-1) <------------is this right

when i use box for maximum/minimum

do i put x value as 4 or 1? (for the first point)

you put 1, due to the fact that x=1 not 4 =)

 

[addikaye03]

Draw neat sketch of the following graphs
2x^2-4x-6

dx/dy=4x-4

4(x-1) <------------is this right

when i use box for maximum/minimum

do i put x value as 4 or 1? (for the first point)

dx/dy=4x-4 <----- that's supposed to be dy/dx, think of what its saying, change in y with respect to a change in x, since the RHS values are in terms of x this is the correct way of writing it. From here:

When dy/dx=0 ( a stationary point occurs)

4x-4=0

4(x-1)=0 [factorising]

x-1=0 [dividing 0 by 4]

therefore x=1

so when you set up the box you put x=1 as a point ( have a point either side of this) and its dy/dx value is 0.

Do you understand now?