Last year's test
If anyone has done the EC practice test, can you help me with 11b? It says to find a slope for the line found in part A so that this line is tangent to the parabola y=x^2. The line I got in part A was y=9-m(x-3), but I'm a bit confused about where to go from here. I tried setting the two equations equal to each other (so that 9-m(x-3)=x^2) and solving for m, but of course that gave me a slope that was in terms of x, not a numerical value. Any ideas?
posted by Casey Blue at 1:24 PM
5 Comments:
I couldn't figure it out either but I got 9+m(x-3)=y...is taylor form y1-m(x-x1) or y1+m(x-x1)?
This comment has been removed by a blog administrator.
I didn't get this problem either. I also got y=9+m(x-3), and Taylor form is y=y1+m(x-x1).
yeah...i wasn't sure about it either..
i think i got it you know how slope would be y2-y1/x2-x1 well you know that x=3 and y=9 and you use the origin, so the slope would be 3. then you take the inverse reciprocal which gives you m=-1/3. it looked right when i graphed it
Post a Comment
<< Home