Makro
Porcupine
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.
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Types of stationary points (1 Viewer)
- Thread starter Makro
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x3.eddayyeeee
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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.
-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.
Last edited:
Makro
Porcupine
Solution:
a = 3
b = -9
So hmm.
a = 3
b = -9
So hmm.
x3.eddayyeeee
Member
- Joined
- Nov 4, 2008
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- Gender
- Male
- HSC
- 2011
a - 3a = -2a.
Hopefully thats right.
that why my answer seemed weird compared to you guys.
