Types of stationary points (1 Viewer)

[Makro]

The curve y= ax³ + bx² - x + 5 has a point of inflexion at (1, -2). Find the values of a and b. I didn't even attempt this question when we had it for homework. Hopefully someone will answer by tomorrow morning.

 

[jet]

Hopefully thats right.

 

[ayehann]

yeah thats what i got too

 

[x3.eddayyeeee]

Sub in (1,-2) in y= ax<SUP>3</SUP> + bx<SUP>2</SUP> - x + 5
-2 = a(1)^3 + b(1)^2 - 1 + 5
-2 = a + b + 4
-6 = a + b . . . . (1)

y= ax<SUP>3</SUP> + bx<SUP>2</SUP> - x + 5
y'= 3ax^2 + 2bx -1
y''= 6ax +2b

Inflexion when y''=0
0 = 6ax + 2b
0 = 2(3ax + b)
0 = 3ax + b
Since inflexion @ x=1
0 = 3a(1) + b
0 = 3a + b . . . . (2)

Solve simultaneous equations. (2) - (1)
0 - (-6) = 3a - a + b - b
6 = 2a
a = 3
sub a = 2 back into (1)
-6 = a + b
-6 = 3 + b
b= -9</SUP>

EDIT. didnt read properly.

 

[Makro]

Solution:

a = 3
b = -9

So hmm.

 

[x3.eddayyeeee]

Hopefully thats right.

a - 3a = -2a.
that why my answer seemed weird compared to you guys.

 

[tommykins]

 

[jet]

LOL. I can't believe i did that. So yes, a = 3 and b = -9.
Lol. Sorry about that. I don't do maths anymore in school.