Question
The curve y = (k-6)x^2 - 8x + k cuts the x-axis at 2 points and has a minimum point. find the range of values of k.
Answer
There are 2 conditions to satisfy for this question ie
(1) the curve cuts the x-axis at two points ie b^2 - 4ac > 0
You have done this part correctly ie the answer is -2 < k < 8
(2) the curve has a minimum point
y = (k-6)x^2 - 8x + k
dy/dx = 2(k-6)x - 8
d^2y/dx^2 = 2(k-6)
For minimum point, d^2y/dx^2 must be > 0
So, 2(k-6) > 0
k > 6
Alternatively, Since the curve has a minimum point ie a > 0, so k - 6 > 0.
Hence k > 6
Now, combine the two inequalities for the overall answer.
-2 < k < 8 and k > 6.
Hence, the overall answer is 6 < k < 8.