find a and b so that f(x)=ax^2+bx has a relative minimum at the point (-1,7)

Question

I really need to learn how to do this problem. Thanks so much!

Answer 1

To be a relative minimum, f'(x)=0.

 

For the value of f(-1)=7:

 

a(-1)² + b(-1) = 7

a - b = 7                        eq1

 

 

For the derivative equal to zero.  That is   f'(1)=0

 

2ax + b = 0

2a(-1) + b = 0

-2a + b = 0                                   eq2

 

Solve the system for a and b.

 

Substitute eq1 into eq2.

 

-2(7 + b) + b = 0

 

Solving for b,

 

-14 - 2b + b = 0

-14 - b = 0

-14 = b

 

Then,

 

a = 7 + b
a = 7 - 14

a = -7

 

 

However, you value of a should be positive.  Since you have a quadratic function, this parabola would have to open upward to have a minimum point. Thus, the leading coefficient must be positive.  Are you sure the point is not a relative maximum?

Answer 2

f(x) = ax² + bx

f'(x) = 2ax + b

f'(-1) = 0 = 2a(-1) + b

f(-1) = 7 = a(-1)² + b(-1)

(i)   -2a + b = 0
(ii)    a - b = 7

Solve (i) and (ii) for a and b and get

 

a = -7

b = -14

 

f(x) = -7x² - 14x

 

But this is a concave down parabola, so the point (-1,7) is a MAXIMUM, not a minimum.  Is this problem worded correctly?