Find constants a,b and c such that the graph of f(x)=x^3+ax^2+bx+c will increase to the point (-3,18), decrease to the point (1,-14) and then continue increasing.

Answer 1

Find constants a,b and c such that the graph of f(x)=x^3+ax^2+bx+c will increase to the point (-3,18), decrease to the point (1,-14) and then continue increasing.
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The derivative is zero at x = -3 and x = 1
f' = 3 x^2 + 2ax + b
0 = 3(9) +2a(-3) +b = 27-6a+b
and
0 = 3(1) +2a(1) +b = 3 +2a+b
so solve those two equations for a and b
then use the value of y = -1 at x = 1 to get c

Answer 2

Thank you Damon, but I still do not understand.

How do I solve for a, if I rearrange 27-6a+b in terms of a, I get a= 27+b/6.
But what do I plug in for b?

Answer 3

f(x)=ax^2+bx+c. The function f is given by where a, b and c are constants.
Top point is (-4,10) and f(8)=2
How?

Answer 4

Hi,
So taking the derivative of the original you get an equation you can use to find a:
f(x)=x^3+ax^2+bx+c
becomes
f'(x)=3x^2+2az+b

Now we know that there are two extremes, one at (-3,14) and another at (1,-14). We also know that at extremes f'(x)=0 therefore we can create two equations to solve for a:
f'(-3)=0=3(-3)^2+2a(-3)+b
simplify and extract b
b=-27+6a

repeat with the x value of 1
f'(1)=0=3(1)^2+2a(1)+b
simplify and extract b
b=-2a-3

now since both the equations are simplified to b we can set them equal to each other and solve for a:
-2a-3=-27+6a
which simplifies to
a=3

then take the two original equations and rearrange them so that b is isolated. This will get you b=-9 (I'm not gonna type it out tho, it's the same process used to achieve the value of a).

Now f(x)=x^3+3x^2-9x+c

Substitute in (1,-14) to the equation f(x) and solve for c:
-14=(1)^3+3(1)^2-9(1)+c
-14=-5+c
c=-9