Asked by George
1. For what values of a will the straight line y=ax+1 have ONE intersection with the parabola y=-x^2-x-8?
Answers
Answered by
Steve
the line intersects where
-x^2-x-8 = ax+1
x^2 + (a+1)x + 9 = 0
In order for there to be a single intersection, the discriminant must be zero. That is,
(a+1)^2 - 36 = 0
See what you can do with that.
http://www.wolframalpha.com/input/?i=plot+-x%5E2-x-8,+y%3D5x%2B1,+y%3D-7x%2B1,+-5%3C%3Dx%3C%3D5
-x^2-x-8 = ax+1
x^2 + (a+1)x + 9 = 0
In order for there to be a single intersection, the discriminant must be zero. That is,
(a+1)^2 - 36 = 0
See what you can do with that.
http://www.wolframalpha.com/input/?i=plot+-x%5E2-x-8,+y%3D5x%2B1,+y%3D-7x%2B1,+-5%3C%3Dx%3C%3D5
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