A curve with equation y = ax^2 + bx + c crossed the x-axis at (-4,0) and (0,0) and also passes through the point (2,120). Where does the curve cross the y-axis?

Printable View

- Mar 20th 2009, 01:21 AMstruckFunctions and graphs question..
A curve with equation y = ax^2 + bx + c crossed the x-axis at (-4,0) and (0,0) and also passes through the point (2,120). Where does the curve cross the y-axis?

- Mar 20th 2009, 01:25 AMProve It
- Mar 20th 2009, 01:55 AMstruck
I knew that and I posted the question wrong :( .. It should have been (0,9) not (0,0) . It just didn't appear to me that the equation should be made of type a(x - s)(x - r) (or similar) .. I solved it though :)

Thanks for the reply. - Mar 20th 2009, 04:06 AMstapel
You have been given three points, being three sets of values for $\displaystyle x$ and $\displaystyle y$. Plug these values into the given equation:

. . . . .$\displaystyle 0\, =\, 16a\, -\, 4b\, +\, c$

. . . . .$\displaystyle 9\, =\, c$

. . . . .$\displaystyle 120\, =\, 4a\, +\, 2b\, +\, c$

Solve the system of equations. (Wink)