# Slopes of curves w/o equation?

I have three points on a Y= -ax^2 + bx + c graph (negative parabola). I don't know how to find the equation with this information. Tangent lines? Help.

Hi donlin. Welcome to PF!
Let's say one of the points is (x0, y0). What's -ax02+bx0+c equal to?

Homework Helper
Your equation is a function of one variable, Y(x)=-ax^2+bx+c. If a point (x0, y0) lies on the parabola, it must satisfy the equation f(x0) = y0. Since you have three points, after 'plugging' every one of them into the equation, you'll have three equations with three unknowns a, b and c, which are the coefficients you need.

I think I have the slope, by connecting point A to B and then rise over run, but I don't know how to get the rest of the equation. Basically, I have a point A and slope, but I don't have the coefficients or y-intercept.

neutrino said:
Hi donlin. Welcome to PF!
Let's say one of the points is (x0, y0). What's -ax02+bx0+c equal to?

I think it's equal to y0.

donlin said:
I think I have the slope, by connecting point A to B and then rise over run, but I don't know how to get the rest of the equation. Basically, I have a point A and slope, but I don't have the coefficients or y-intercept.
You can't get the slope at one point of a parabola by connecting two points on it. It's different at every point. Use the method that radou posted.

HallsofIvy