# Math Help - Polynomial Help Please Urgent!!!

1. ## Polynomial Help Please Urgent!!!

The question is: Sketch a polynomial function of degree 3 that has x-intercepts 1, 2, and 3, has a y intercept of 6, and passes through the point (-1,25). Can you actually have the graph you just drew?

So, i tried making the formula for the function and i started out with the base (x-1)(x-2)(x-3) to get the x intercepts. Then i added 6 to the end to bump the function up to have a y intercept of 6. Finally, i plugged in (-1,25) to the function in order to find the leading coefficiant of a(x-1)(x-2)(x-3)+6=f(x)

I then solve for A and it comes out something like .7196666 repeating. I check the graph F(x)=.719666666(x-1)(x-2)(x-3)+6 and it looks nothing like its supposed to. Please Help. THANKS!!!!

2. ## ...a push in the right direction

Hi Aussiekid90:

The general form of a cubic polynomial function is f(x)=ax^3+bx^2+cx+d. Recall that a "polynomial of degree n" is, by definition, the sum of all non-negative integral powers of x (meaning with integer exponents), having "greatest" degree, n, and all coefficients real.

Now, you are given that each of (1,0),(2,0),(3,0),(0,6)and(-1,25)is a solution of the above noted function. That is a mountain of information. First, the value of constant d is a gift. Next, using said mountain, write two equations of the form 0 = ..., such that the terms comprising the "..." contain only a, b, c and 6 (woops, I slipped!). Having done so, you will likely observe a pretty nifty occurance, should you happen to possess sufficient insight to consider adding these equations, thus constructing yet another in the process.

I am confident that, having arrived at the juncture marked by the smiley above, the rest will just "sorta fizzle all together" (in the words of a really cool dude who answers to the name "Neil Young").

Happy function hunting,

Regards,

Rich B.

3. Originally Posted by aussiekid90
The question is: Sketch a polynomial function of degree 3 that has x-intercepts 1, 2, and 3, has a y intercept of 6, and passes through the point (-1,25). Can you actually have the graph you just drew?

So, i tried making the formula for the function and i started out with the base (x-1)(x-2)(x-3) to get the x intercepts. Then i added 6 to the end to bump the function up to have a y intercept of 6. Finally, i plugged in (-1,25) to the function in order to find the leading coefficiant of a(x-1)(x-2)(x-3)+6=f(x)

I then solve for A and it comes out something like .7196666 repeating. I check the graph F(x)=.719666666(x-1)(x-2)(x-3)+6 and it looks nothing like its supposed to. Please Help. THANKS!!!!
As you said cubic polynomial thus $a(x-1)(x-2)(x-3)$ where $a$ is a non-zero real number. The problem states that the y-intercept is 6, thus when $x=0$ then $f(x)=6$ thus,
$a(-1)(-2)(-3)=6$ thus, $a=-1$. This is it.
But the problem is when you said $(-1,25)$ it is impossible however, you probably meant $(-1,24)$ then it is good.