# Functions and graphs question..

• Mar 20th 2009, 02:21 AM
struck
Functions 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, 02:25 AM
Prove It
Quote:

Originally Posted by struck
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?

The y-intercept occurs where x = 0.

You're told that the curve passes through the origin.

Therefore the y-intercept must be 0.
• Mar 20th 2009, 02:55 AM
struck
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 :)

• Mar 20th 2009, 05:06 AM
stapel
Quote:

Originally Posted by struck
A curve with equation y = ax^2 + bx + c crossed the x-axis at (-4,0) and (0,9) and also passes through the point (2,120). Where does the curve cross the y-axis?

You have been given three points, being three sets of values for $x$ and $y$. Plug these values into the given equation:

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

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

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

Solve the system of equations. (Wink)