How to find the point of intersection for all values of m,c and a?

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Can someone please explain to me how to do question 22? Thanks!

1 Answer
Jun 15, 2018

The answer is (D). Set the two curves equal and solve for #x# to find points of intersection, then deduce the parameter conditions for existence of said points.

Explanation:

To find the points of intersection of the straight line #y=mx+c# and the parabola #y=ax^2#, set their #y# values equal and solve for #x#:

#ax^2=mx+c#
#ax^2-mx-c=0#
#x=(m+-sqrt(m^2+4ac))/(2a)#

We see from this that the curves do not intersect when the discriminant #m^2+4ac<0#. In this case the roots of the quadratic have an imaginary component.

Rearrange this condition:
#m^2+4ac<0#
#4ac<-m^2#

If #a>0#, then
#c<-m^2/(4a)#
If #a<0#, then
#c> -m^2/(4a)#

So the answer is (D). Your written working on the photo shows that you're very nearly there already; you just need to note that when #a# is negative, dividing the inequality through by it changes the inequality direction.
NB There are two parameter regions that work, but only one is listed as an option in the question - they aren't looking for the complete solution of all possible regions, rather for you to say which of their options is part of the complete solution.