On Tue, 12 Feb 2008 04:26:47 -0800 (PST), Dougsd1r
<dougsdir@[EMAIL PROTECTED]
> wrote in
<news:4f3823b6-6c8d-4272-8439-531af7078b81@[EMAIL PROTECTED]
>
in alt.algebra.help:
> aaarrgghh , I'm still stuck
To make this self-contained, I'm inserting the original
problem here:
A parabola has equation: y = ax^2 + bx + c.
It p***** through the origin, and the tangent at (1,0)
on it makes an angle of 45 degrees with OX.
Find the actual equation on the parabola
> I know that 0 and 1 are roots
That tells you that x and x - 1 are factors of the
quadratic, which you already know is of the form ax^2 + bx
(since you know that c = 0). In other words, you know that
ax^2 + bx is a constant multiple of x(x - 1), say kx(x - 1).
Now kx(x - 1) = kx^2 - kx, so the only possibility is that
b = -a in the original expression, which then becomes y =
ax^2 - ax. (This purely algebraic fact is what you were
missing, I think. More generally, you want to remember that
if p(x) is a polynomial, then r is a root of p(x) -- i.e.,
p(r) = 0 -- if and only if x - r is a factor of p(x).)
Now all that remains is to find the coefficient a, and
that's what the information about the tangent line at (1, 0)
is for.
> so if the gradient of the tangent at (1,0) is 1 then the
> gradient of the tangent at (0,0) is -1
The term 'gradient' has a technical meaning that isn't
really appropriate here; you want simply 'slope'.
> I can calculate the equations of these 2 lines:
> y = -x and y = x - 1
You don't need the first of these, or anything that
followed. Go back to the equation of the parabola as it now
stands, y = ax^2 - ax, and use it to calculate the slope of
the tangent line at x = 1. You should get a value expressed
in terms of the coefficient a; and since you know that the
correct value is 1, you'll be able to determine that a = 1.
[...]
Brian
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