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# Crossing Y Axis watch

1. A curve with equation y=ax^2+bx+c crosses the x axis at ( -4 , 0 ) and ( 9, 0 ) and also passes through the point ( 1, 120 ). Where does it cross the Y axis ?

And for another question, say when I have 3 lines to find a common point of intersection, do I just do lines 1 and 2 seperately. Then 2 and 3 ? And find common points ?
2. The y-axis is where the x-coordinate is zero. So set x equal to zero to find what the y-coordinate is, and then this is where it crosses the y-axis. If you're stuck on finding a,b,c, then notice that it x=-4 and x=9 are the roots of the equation (since that's where it crosses the x-axis), so it must be in the form . Sub in the coordinates of the extra point to find the value of .

Two non-parallel lines (in 2D space) only ever intersect at one point, so you need to:
(a) Find where two of the lines intersect
(b) Check that the third line intersects either of the other two lines at the same point (equivalently, just show that the point of intersection of the other two lines also lies on the third line)
3. Okay thanks, 2nd question should be ok, but I'm not sure what you mean by setting it to zero though, which one do I set to zero out of

( -4 , 0 ) and ( 9, 0 ) ( 1, 120 ) ?
4. (Original post by Gelato)
Okay thanks, 2nd question should be ok, but I'm not sure what you mean by setting it to zero though, which one do I set to zero out of

( -4 , 0 ) and ( 9, 0 ) and also passes through the point ( 1, 120 ) ?
What?

First you find a,b,c. Then your equation will be something like (this is just an example, not the answer to your question). You then set x equal to zero in this equation -- when x is equal to zero, the curve is crossing the y-axis (draw a diagram to see why). Setting x equal to zero then gives the y-coordinate of this point; in this case, (5,0).
5. I think we did this stuff in class today..
i slept through the lesson

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