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Can someone help me find a and b of a function when you are given an inflection point?

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Find a and b such that the function f(x)=ax^3+bx^2 has an inflection point at (−3,216) 

a = ? 
b = ?

asked Nov 12, 2014 in PRECALCULUS by anonymous

1 Answer

0 votes

The function f(x)=ax³+bx² .

The curve passes through (−3,216) .

So f(-3)=a(-3)³+b(-3)²

f(-3) = -27a+9b 

216 = -27a+9b --------->(1)

x = -3 is a point of inflection, so f''(-3) = 0 .

f(x)=ax³+bx²

f '(x) = 3ax² + 2bx 

f ''(x) = 3(2)ax + 2b

f ''(x) = 6ax + 2b

f ''(-3) = 6a(-3) + 2b

f ''(-3) = -18a + 2b

0 = -18a + 2b

-18a + 2b = 0 

-2(9a - b) = 0  

9a - b = 0

9a = b 

Put b = 9a in equation (1)

216 = -27a+9(9a)

216 = -27a+81a

216 = 54a

a = 216/54

a = 4 .

Now substitute a = 4  in  b = 9a .

b = 9(4)

b = 36

So the values are a = 4 and b = 36 .

answered Nov 12, 2014 by yamin_math Mentor

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