# Curve Sketching

• May 26th 2010, 06:02 PM
steezin
Curve Sketching
Let f(x) = ax3 + bx2 + cx + 1. Determine the values of a,b, and c so that f(x) has a point of inflection at x = 2, a local minimum at x = -2, and f(1) = 2.

So, f'(x) = 3ax2 + 2bx + c
f"(x) = 6ax + 2b

I know that f'(x) = 0 when x = -2 so: 12a - 4b + c = 0
and f"(x) = 0 when x = 2 so: 6a + b = 0
and a + b + c = 1 since f(1) = 2

I'm not sure if i'm going in the right direction with this question but i'm stuck at this point. Any help would be appreciated.
• May 26th 2010, 06:20 PM
skeeter
Quote:

Originally Posted by steezin
Let f(x) = ax3 + bx2 + cx + 1. Determine the values of a,b, and c so that f(x) has a point of inflection at x = 2, a local minimum at x = -2, and f(1) = 2.

So, f'(x) = 3ax2 + 2bx + c
f"(x) = 6ax + 2b

I know that f'(x) = 0 when x = -2 so: 12a - 4b + c = 0
and f"(x) = 0 when x = 2 so: 6a + b = 0
and a + b + c = 1 since f(1) = 2

I'm not sure if i'm going in the right direction with this question but i'm stuck at this point. Any help would be appreciated.

solve the system of equations ...

$12a-4b+c=0
$

$6a+b=0$

$a+b+c=1$

subtract the 3rd equation from the 1st ...

$11a-5b=-1$

from the 2nd equation ... $b = -6a$

$11a-5(-6a) = -1$

$a = -\frac{1}{41}$

$b = \frac{6}{41}$

$c = \frac{36}{41}$

yes, I know ... some ugly coefficients.