# Thread: finding constants of function

1. ## finding constants of function

The function f(x)=ax^3+bx^2+cx+d has turning point (2, -4) and cuts the y axis at (0, 2). Find constants a, b, c, and d.

can someone check my work please?

a=0
b=3/2
c=-6
d=2

2. if you have 2 equations and 4 constants, certainly you'll find more than 1 solution.

how did you get to that answer?

edit:I notice now about the 'turning' point. I dont know if s some mathematical term in English that provides more info.

3. what's that supposed to mean?

4. it means I dont know what turning means

if its something like $\frac{df(x)}{dx}=0$, you get an extra euqation. its still not enough to provide only one solution.

but if it means that $\frac{df(x)}{dx}=0$ and $\frac{d^2f(x)}{dx^2}=0$ (this is only an example, as I said I dont know what it means) you get for equations and 4 constants you can have a system and solve it for only one solution.

if it doesnt mean nothing you still have a system but you'll get more than 1 solution.

5. Originally Posted by Haytham
it means I dont know what turning means

if its something like $\frac{df(x)}{dx}=0$, you get an extra euqation. its still not enough to provide only one solution.

but if it means that $\frac{df(x)}{dx}=0$ and $\frac{d^2f(x)}{dx^2}=0$ (this is only an example, as I said I dont know what it means) you get for equations and 4 constants you can have a system and solve it for only one solution.

if it doesnt mean nothing you still have a system but you'll get more than 1 solution.
oh, well my teacher meant by max/min point = turning point.. so I took it as f'(x)

Can you check my answers with that being said now?

6. considering that, and only that, your solution is incomplete. that is only one solution, you have to find all.

with the f(x)' info you get 3 equations

$2=d$

$-4=8a+4b+2c+d$

$0=12a+4b+c$

you need to solve those 3. you will get something like
$d=2$

$c=4a-6$

$b=\frac{3}{2}-4a$

$a \epsilon R$

this means, you can choose any value for a, and calculate the values of b,c and d, with that a and get different solutions.

7. Originally Posted by skeske1234

d=2
This is correct

Originally Posted by skeske1234

a=0
b=3/2
c=-6
How did you get these values?

8. Here is my work, please check:

f'(x)=0 when
0=12a+4b+c

f(2)=-4
-4=24a+4b+2c+d

2=d

-12a-4b=c

-4=24a+4b+2(-12a-2b)+2
3/2=b
a=0

-4(3/2)=c
6=c

Is this ok?

9. Originally Posted by skeske1234
The function f(x)=ax^3+bx^2+cx+d has turning point (2, -4) and cuts the y axis at (0, 2). Find constants a, b, c, and d.
This gives the following

$f(2) = -4$

$f(0) = 2 \Rightarrow d = 2$

$f'(2)=0$

Originally Posted by skeske1234

f'(x)=0 when
0=12a+4b+c

You mean $f'(2)=0 \Rightarrow 0=12a+4b+c$

Originally Posted by skeske1234

f(2)=-4
-4=24a+4b+2c+d

Given $f(x) = ax^3+bx^2+cx+2$

$f(2) = 2^3a+2^2b+2c+2$

$-4 = 2^3a+2^2b+2c+2$

$-4 = 8a+4b+2c+2$