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
it means I dont know what turning means
if its something like $\displaystyle \frac{df(x)}{dx}=0$, you get an extra euqation. its still not enough to provide only one solution.
but if it means that $\displaystyle \frac{df(x)}{dx}=0$ and $\displaystyle \frac{d^2f(x)}{dx^2}=0$ (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.
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
$\displaystyle 2=d$
$\displaystyle -4=8a+4b+2c+d$
$\displaystyle 0=12a+4b+c$
you need to solve those 3. you will get something like
$\displaystyle d=2$
$\displaystyle c=4a-6$
$\displaystyle b=\frac{3}{2}-4a$
$\displaystyle a \epsilon R$
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.
This gives the following
$\displaystyle f(2) = -4$
$\displaystyle f(0) = 2 \Rightarrow d = 2$
$\displaystyle f'(2)=0$
You mean $\displaystyle f'(2)=0 \Rightarrow 0=12a+4b+c$
Given $\displaystyle f(x) = ax^3+bx^2+cx+2 $
$\displaystyle f(2) = 2^3a+2^2b+2c+2 $
$\displaystyle -4 = 2^3a+2^2b+2c+2 $
$\displaystyle -4 = 8a+4b+2c+2 $