On Tue, 12 Feb 2008, Dougsd1r wrote:
> Here is the question
>
> A parabola has equation: y = ax^2 + bx + c.
>
> It p***** through the origin,
Thus y(0) = 0 and c = ??
> and the tangent at (1,0) on it makes an
> angle of 45 degrees with OX.
The tangent at (1,0) ? But for there to be a tangent at (1,0)
what does that mean? That it has to pass through (1,0), does it not?
Thus produce a second equation involving a,b and c. What do you get?
> Find the actual equation on the parabola
>
> So far i have
>
> Since the parabola p***** through the origin then c = 0
>
> so the equation is just y = ax^2 + bx
>
> Also the tangent of the line at (1,0) is 1 (since tan 45 = 1)
> i dont know if this is needed or not but the equation of this line is
Correct spelling is "I don't"
> therefore
> y = x - 1
>
Show your work.
> Now this is as far as I got. This question comes at the end of an
> introduction to Differentiation chapter the answer in the book is y =
> x^2 + x Ive tried to use the answer to work backwards but not getting
> anywhere
>
If a curve y = f(x) goes throught a point (x0,y0), then
the slope of the tangent to the curve at the point (x0,y0)
is y'(x0) = f'(x0,y0), or dy(x0)/dx. That's what you need
for a third equation for a,b,c.
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