fing

[red and white kop!]

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?

 

[Deleted member 4993]

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?

The f(t) = 0 at x=p and x=q

so

p = 3 and q = 1 (or p = 1 and q =3) ? now find 'a' (one equation - one unknown) - too much party last night????

 

[BigGlenntheHeavy]

f(t) = a(t-p)(t-q)^2

f(1) = 0 =a(1-p)(1-q)^2

f(3) = 0 = a(3-p)(3-q)^2

f(0) = -4.5 = a(-p)(-q)^2 = -apq^2 or 4.5 = apq^2

Now, just by inspection for f(1)=0 and f(3) = 0, both equations are satisfied when p=1 and q=3 or when p=3 and q=1.

Ergo, when p-1 and q=3, a = 1/2 (4.5 = apq^2), and when p=3 and q=1, a= 3/2.

Hence, combining terms, we get two cubic equations, viz.,

f(t) = (1/2)[t^3-7t^2+15t-9] and g(t) = (3/2)[t^3-5t^2+7t-3].

It behooves one to check that these two equations satisfied the 3 points (which they do).

Note: When p=1, q can't equal one and a can't equal 0 and when q=1, p can't equal 1 and a can't equal 0, why?

Additional note: red and white kop!, who is or was Chelsea? A lover that jilted you. Get over it, as there is more than one fish in the sea.