Differentiation again. (1 Viewer)

[Crosswinds]

Okie another stupid (probably) question:

"The curve y = ax^3 + bx^2 - x +5 has a point of inflexion at (1, - 2). Find the values of a and b."

I suppose you have get two equations and solve them simultaneously or something but I can only get one and I don't know what else to do.

Thanks.

 

[Timothy.Siu]

Okie another stupid (probably) question:

"The curve y = ax^3 + bx^2 - x +5 has a point of inflexion at (1, - 2). Find the values of a and b."

I suppose you have get two equations and solve them simultaneously or something but I can only get one and I don't know what else to do.

Thanks.

y'=3ax^2+2bx-1
0=3a+2b-1
y''=6ax+2b
0=6a+2b

subtracting them
3a+1=0
a=-1/3
b=1

 

[lolokay]

your other equation comes from the fact that y=-2 when x=1

 

[random-1005]

Okie another stupid (probably) question:

"The curve y = ax^3 + bx^2 - x +5 has a point of inflexion at (1, - 2). Find the values of a and b."

I suppose you have get two equations and solve them simultaneously or something but I can only get one and I don't know what else to do.

Thanks.

(1,-2) LIES ON THE CURVE (IE Y)
therefore subing into y equation
-2=a+b-1+5
a+b=-6 ...........eqn 1 rearrange a=-6-b eqn 2

inflexions occur when y'' second derivative =0
y'=3ax^2+2bx-1
y''= 6ax+2b (subing in 1,-2)
-2=6a+2b..... eqn 3

eqn2 into eqn 3
-2=-36-6b+2b
-4b=34
b=-8.5

sub into eqn 2
a=-6-(-8.5)= 2.5
therefore a= 2.5 and b= -8.5

 

[Tully B.]

y'=3ax^2+2bx-1
0=3a+2b-1
y''=6ax+2b
0=6a+2b

subtracting them
3a+1=0
a=-1/3
b=1

That's assuming that the point is a horizontal point of inflexion.

 

[random-1005]

y'=3ax^2+2bx-1
0=3a+2b-1
y''=6ax+2b
0=6a+2b

subtracting them
3a+1=0
a=-1/3
b=1

why are subing conditions into y'?
(1,-2) is a point on the curve and an inflexion, not a stationary pt

 

[Tully B.]

(1,-2) LIES ON THE CURVE (IE Y)
therefore subing into y equation
-2=a+b-1+5
a+b=-6 ...........eqn 1 rearrange a=-6-b eqn 2

inflexions occur when y'' second derivative =0
y'=3ax^2+2bx-1
y''= 6ax+2b (subing in 1,-2)
-2=6a+2b..... eqn 3

eqn2 into eqn 3
-2=-36-6b+2b
-4b=34
b=-8.5

sub into eqn 2
a=-6-(-8.5)= 2.5
therefore a= 2.5 and b= -8.5

In equation 3, 6a +2b = 0 (not -2)
This is because at (1, -2), y''= 0, since it is a point of inflexion.

 

[random-1005]

In equation 3, 6a +2b = 0 (not -2)
This is because at (1, -2), y''= 0, since it is a point of inflexion.

ohh yes, its ment to be when x=1, y''=0
thx

 

[Crosswinds]

Ohhh okay get it now. Thankyou!

 

[smp211]

here is how i did it :

y = ax^3 + bx^2 - x + 5
y' = 3ax^2 +2bx - 1
y'' = 6ax + 2b

pt of infl;
.:. 0 = 6a(1) + 2b
.:. 0 = 6a + 2b [1]

we know a pt on the curve is (1, -2)
.:. -2 = a(1)^3 + b(1)^2 - 1 + 5
.:. -2 = a + b -4
.:. -6 = a + b [2]

solve simultaneously:
-36 = 6a + 6b [3]

[3] - [1]
.:. -36 = 4b
.:. b = -9
sub in value of b into [2]
.:. a = 3

so to answer the question, a = 3 & b = -9

am i right?

 

[Crosswinds]

Yep, 3 and -9 were the answers.