# Thread: Manipulating cubic functions: x-intercepts

1. ## Manipulating cubic functions: x-intercepts

Hi,

I'm trying to create a cubic function that has only one x-intercept of -5 and a y-intercept of around 10.
The y-intercept doesn't have to be exact.
I can't seem to get this as I dont know what to do with the roots.
I get to:
(x+5)(x+-a)(x+-b)
and i dont know what to do from there.
Any help is greatly appreciated.

Thanks,
John

2. Hello, John!

I'm trying to create a cubic function that has only one x-intercept of -5
and a y-intercept of around 10.
The y-intercept doesn't have to be exact.

Since $\displaystyle x=-5$ is a zero of the cubic, $\displaystyle (x+5)$ is a factor.

Then the cubic is the product of $\displaystyle (x+5)$ and a quadratic.
. . $\displaystyle f(x) \;=\;(x+5)(ax^2+bx + c)$

The $\displaystyle y$-intercept will be 10 if $\displaystyle c = 2.$
. . $\displaystyle f(x) \;=\;(x+5)(ax^2 + bx + 2)$

Since $\displaystyle x = -5$ is the only $\displaystyle x$-intercept, the quadratic has no real roots.

. . That is: .$\displaystyle b^2 - 4ac \:<\:0$

We have: .$\displaystyle b^2 - 8a \:<\:0 \quad\Rightarrow\quad a \:>\:\frac{b^2}{8}$

Let $\displaystyle b = 2 \quad\Rightarrow\quad a = 1$

And we have: .$\displaystyle f(x) \:=\:(x+5)(x^2 + 2x + 20 \:=\:x^3 + 7x^2 + 12x + 10$

. . . . . one of a brizillion possible cubics.

3. Oh, I understand now!
Thanks heaps.
although, where does $\displaystyle b^2 - 4ac \:<\:0$ come from?