9514 1404 393
Answer:
a < -5 or 11 < a
Step-by-step explanation:
For there to be two points of intersection, there must be 2 real solutions to ...
ax +9 = -2x^2 +3x +1
Rewriting to standard form, we have ...
2x^2 +(a -3)x +8 = 0
The discriminant of this equation is ...
B^2 -4AC = (a -3)^2 -4(2)(8) = a^2 -6a -55
We want this to be greater than 0. Factoring, we get ...
(a -11)(a +5) > 0
This will be true for values of a that are less than -5 or greater than 11.
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The attached graph shows lines and their points of intersection for a=-6 and a=12. It is not hard to see that for slopes less steep than these values, the lines with a y-intercept of 9 may not intersect the parabola.
