right track? (1 Viewer)

[darshil]

curve is 3x^2-x^3

question: find the equation of a tangent to he curve at point R(-1,4)

1) get first derivative f'(x)=6x-3x^2
2) sub in x value of given point R which is -1
3) f'(-1) = -9
4)so at the point R the gradient is -9
5) use (y-y1)=m(x-x1) to get the equation of the tangent
6) (y-4)=-9(x+1)
7) which = y-4=-9x-9 (which can be made into y=-9x+5
is this it? is it correct? Thanks for the help!

 

[Drongoski]

curve is 3x^2-x^3

question: find the equation of a tangent to he curve at point R(-1,4)

1) get first derivative f'(x)=6x-3x^2
2) sub in x value of given point R which is -1
3) f'(-1) = -9
4)so at the point R the gradient is -9
5) use (y-y1)=m(x-x1) to get the equation of the tangent
6) (y-4)=-9(x+1)
7) which = y-4=-9x-9 (which can be made into y=-9x+5
is this it? is it correct? Thanks for the help!

Yes. Congrats!

 

[darshil]

Thank you, i really appreciate it !

 

[Drongoski]

Thank you, i really appreciate it !

Is darshil yr actual name?

 

[darshil]

yeah man, darshil shah

 

[Drongoski]

yeah man, darshil shah

i thought it's prob a punjabi sikh name, e.g. darshan singh bachan frinstance. just curious.

 

[darshil]

hahah nah man, I'm from Gujurat.

 

[study-freak]

curve is 3x^2-x^3

question: find the equation of a tangent to he curve at point R(-1,4)

1) get first derivative f'(x)=6x-3x^2
2) sub in x value of given point R which is -1
3) f'(-1) = -9
4)so at the point R the gradient is -9
5) use (y-y1)=m(x-x1) to get the equation of the tangent
6) (y-4)=-9(x+1)
7) which = y-4=-9x-9 (which can be made into y=-9x+5
is this it? is it correct? Thanks for the help!

There is a mistake at exactly where you highlighted. y-4=-9x-9 is y=-9x-5.
Otherwise, correct solution.

 

[darshil]

OH thanks man!

 

[Drongoski]

OH thanks man!

darshil

Kem cho ?

I'm so sorry I did not spot the "-5" mistake referred to by study-freak.

 

[darshil]

haha no worries

Bas Majama (= Thamhe?