Cubic Functions Derivatives

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May 14th 2008, 02:53 AM
andrew2322

Cubic Functions Derivatives

Hello all, i need assistance with these, thanks

1.) Find a cubic function whose graph touches the x-axis at x = -4, passes through the origin and has a value of 10 when x = 5.

2.) The function f(x) = 2x^3 + ax^2 - bx +3. When f(x) is divided by x - 2 the remainder is 15 and f(1) = 0

A.) Calculate the values of a and b
B.) Find the other two factors of f(x)

3.) The polynomial P(x) = x^3 + ax^2 + bx - 9 has zeros at x = 1 and x = -3. The values of a and b are?

Please guys, full working out and answer, me be stupid :P
this isnt homework, its revision so you dont have to worry about me cheating. THANKS!
May 14th 2008, 04:36 AM
earboth

Quote:

Originally Posted by andrew2322

Hello all, i need assistance with these:

1.) Find a cubic function whose graph touches the x-axis at x = -4, passes through the origin and has a value of 10 when x = 5.
...

The general equation of a cubic function is:

$f(x)=ax^3+bx^2+cx+d$ and it's first derivative

$f'(x)=3ax^2+2bx+c$

You already know:

$\begin{array}{l}f(0)=0 \\ f(-4)=0 \\ f'(-4)=0 \\ f(5)=10\end{array}$ ...... $~\Longrightarrow~$ ...... $\begin{array}{l}d=0 \\ -64a+16b-4c =0 \\ 48a - 8b + c=0 \\ 125a+25b+5c=10\end{array}$

Solve the system of simultaneous equations.
May 14th 2008, 04:51 AM
earboth

Quote:

Originally Posted by andrew2322

Hello all, i need assistance with these...

2.) The function f(x) = 2x^3 + ax^2 - bx +3. When f(x) is divided by x - 2 the remainder is 15 and f(1) = 0

A.) Calculate the values of a and b
B.) Find the other two factors of f(x)

...

to A.):

From f(1) = 0 you know that

$a-b+5=0$ ...... [1]

Use long divison to calculate the remainder. I've got as remainder r = 19 +4a - 2b which is according to your problem 15:

$19+4a-2b = 15$ ...... [2]

Solve this system of simultaneous equations for a and b. I've got a = 3 and b = 8

Thus the equation of the function becomes:

$f(x)=2x^2+3x^2-8x+3$

to B.):

Use synthetic division :

$(2x^2+3x^2-8x+3) \div (x-1) = 2x^2+5x-3$

Now solve the quadratic equation

$2x^2+5x-3 = 0$

You should get $x = \frac12~\vee~x=-3$ and therefore the other 2 factors are:

$(x+3)\left(x-\frac12\right)$
May 15th 2008, 02:54 AM
andrew2322

hey

im so confused lol :(