Question 1081227
We have two equations and two unknowns:
y = x^2 + 5x - 7
y = ax^2 + (a+4)x + 9
Setting the y values equal, we obtain a quadratic equation in x:
x^2 + 5x - 7 = ax^2 + (a+4)x + 9 -> (1-a)x^2 + (1-a)x - 16 = 0 [1]
The number of solutions of a quadratic depends on the value of the discriminant, b^2 - 4ac
If b^2 - 4ac < 0 there are no solutions
If b^2 - 4ac = 0 there is one solution
If b^2 - 4ac > 0 there are two solutions
For simplicity, set k = 1-a
kx^2 + kx - 16 = 0
b^2 - 4ac = k^2 + 64k
a) k^2 + 64k < 0
This will be true if k<0 and k>-64 -> 1-a<0 -> a>1 AND 1-a>-64 -> a<65
b) k^2 + 64k = 0
k(k+64) = 0 
This has two solutions, k=0 and k=-64, -> a=1 and a=65
But a=1 cannot yield a solution, since from [1], we would have -16 = 0, which is obviously false.
Therefore, a=65 is the only value for a which yields one intersection point
c) can be solved in a similar manner