1. ## whopping equation

values of s along y axis
values of t along x axis
s=f(t)=a (t-p)(t-q)^2
where a, p, q are constants
the curve meets the coordinate axes at the points (0, -4.5) (1, 0) (3, 0)
find values for a, p ,q

i came up with three equations but they are so massive i doubt im on the right track.
i found
0 = a ( 27 - 18q + 3q^2 - 9p + 6pq - pq^2)
0 = a ( 1 - 2q + q^2 - p + 2pq - pq^2)
4.5 = apq^2
are these correct? if so, how can they be solved in a time period shorter than say a month?
somebody please show me step by step?

2. You're thinking too hard.

values of s along y axis
values of t along x axis
s=f(t)=a (t-p)(t-q)^2
where a, p, q are constants
the curve meets the coordinate axes at the points (0, -4.5) (1, 0) (3, 0)
find values for a, p ,q
Sorry, but I'm going to change the variables s & t to x & y.
$\displaystyle y = a(x - p)(x - q)^2$

The 2nd and 3rd points are x-intercepts.
(Note: if you have an equation in factored form, like y = (x - m)(x - n), finding the x-intercepts is really easy; they're the numbers subtracted from x. In my example, they would be m and n.)

That means that p = 1 and q = 3, or p = 3 and q = 1. Plug in each pair, with the 1st point, to find a.

p = 1, q = 3:
\displaystyle \begin{aligned} -4.5 &= a(0 - 1)(0 - 3)^2 \\ -4.5 &= -9a \\ a &= 0.5 \end{aligned}

p = 3, q = 1:
\displaystyle \begin{aligned} -4.5 &= a(0 - 3)(0 - 1)^2 \\ -4.5 &= -3a \\ a &= 1.5 \end{aligned}

So I found two sets of values of a, p, and q. Unless there is more information that I'm missing...?

01

3. Hello everyone

I agree with yeongil except that there's a sign wrong: $\displaystyle p =1, q=3, a = -0.5$ and $\displaystyle p = 3, q = 1, a = -1.5$.

4. but where did i go wrong? could this have been solved my way?

5. furor celtica: perhaps, but it would be nasty. Why do it if the method I presented is easier?

Grandad: I double-checked, and I'm still getting positive values for a.

Come to think of it, a has to be positive. Looking at the original function:
$\displaystyle y = a(x - p)(x - q)^2$

All of the roots of the cubic are known: 1 real root, and 1 double real root. To the left of the roots we have a known point below the x-axis: (0, -4.5). Our cubic, then, is such that
as x -> -∞, y -> -∞ and
as x -> ∞, y -> ∞.
a, the leading coefficient, must therefore be positive.

01

6. Originally Posted by yeongil
Grandad: I double-checked, and I'm still getting positive values for a.
You're quite right - sorry!