1. ## polynomial caculus

Determine a,b,c and d so that the graph of

$y= ax^3+bx^2+cx+d$

has a point of inflection at the origin and a relative maximum at the point (2,4)

2. $y = ax^3 + bx^2 + cx + d$

so $\frac{dy}{dx} = 3ax^2 + 2bx + c$

and $\frac{d^2y}{dx^2} = 6ax + 2b$

For a point of inflection, $\frac{d^2y}{dx^2} = 0.$

So $6ax + 2b = 0$ at $x = 0$ (origin)

$\therefore b = 0$.

Now at (2,4) $\frac{dy}{dx} = 0$ as we have a local maximum, and we must make sure $\frac{d^2y}{dx^2} < 0$ so it is a max and not a min.

Substituting in the values:
$
12a + c = 0$
(b is gone as it equals zero)

$\Rightarrow c = -12a$

Now, we know that $y = ax^3 - 12ax + d$

We know that at $x = 0$, $y = 0$, so you can see $d = 0$.

Now we get $y = ax^3 - 12ax$.

You know that at $x = 2$, $y = 4$. Simply substitute these values into the equation to find a. Then you can find c.

3. Hello, Rimas!

Determine $a,b,c,d$ so that the graph of: . $f(x) \:=\:ax^3+bx^2+cx+d$

has a point of inflection at the origin and a relative maximum at the point (2, 4)
Since the origin (0,0) is on the graph: . $f(0) = 0$
. . We have: . $a\!\cdot0^3 + b\!\cdot\!0^2 + c\!\cdot\!0 + d \:=\:0 \quad\Rightarrow\quad\boxed{d \:=\:0}$
The function is: . $f(x) \:=\:ax^3+bx^2 + cx$

The first derivative is: . $f'(x) \:=\:3ax^2 + 2bx + c$
The second derivative is: . $f''(x) \:=\:6ax + 2b$

Since the origin is an inflection point: . $f''(0) \,=\,0$
. . We have: . $f''(0) \:=\:6a\!\cdot\!0 + 2b \:=\:0\quad\Rightarrow\quad\boxed{b \:=\:0}$
The function is: . $f(x) \:=\:ax^2 + cx$

Since (2, 4) is on the graph: . $f(2) \,=\,4$
. . $f(2) \:=\:8a + 2c \:=\:4 \quad\Rightarrow\quad c \:=\:2 - 4a$ .[1]

Since (2, 4) is a critical point: . $f'(2) \,=\,0$
. . We have: . $f'(2) \:=\:3a\!\cdot\!2^2 + c \:=\:0 \quad\Rightarrow\quad c \:=\:-12a$ .[2]

Equate [1] and [2]: . $2 - 4a \:=\:-12a \quad\Rightarrow\quad\boxed{ a \:=\:\text{-}\tfrac{1}{4}}$

Substitute into [2]: . $c \:=\:-12\left(\text{-}\tfrac{1}{4}\right) \quad\Rightarrow\quad\boxed{ c \:=\:3}$

Therefore: . $f(x) \:=\:-\frac{1}{4}x^3 + 3x$

4. Originally Posted by Rimas
Determine a,b,c and d so that the graph of

$y= ax^3+bx^2+cx+d$

has a point of inflection at the origin and a relative maximum at the point (2,4)
since the curve passes through the origin, $d = 0$.

$y = ax^3 + bx^2 + cx$

curve also passes through the point (2,4) ...

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

$2 = 4a + 2b + c$

$y' = 3ax^2 + 2bx + c$

since y has a relative max at (2,4) ...

$y' = 0$ at $x = 2$

$0 = 12a + 4b + c$

$y'' = 6ax + 2b$

$y'' = 0$ at $x = 0$

$0 = 2b$

$b = 0$

so ... now you have the two equations

$12a + c = 0$

$4a + c = 2$

finish up by solving for a and c.