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.
Since it cuts at 2 points, discriminant should be >0.
So I subbed using the b^2 - 4ac method...
I got -2<k<8. my friend says it should be 6<x<8. where is it that I went wrong? is it the wrong application of method or whatsoever?
the graph has a minimum point, so k must be more than 6, if not x^2 will be negative, which will make the graph to have either a min or max point. .
Hi Anpanman,
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
You have not considered this condition in your answer.
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
Now, combine the two inequalities for the overall answer.
-2 < k < 8 and k > 6.
Hence, the overall answer is 6 < k < 8.
Thank you for your kind attention.
Regards,
ahm97sic
I doubt that the setter intended it to be a quad+diff question. It is simply a basic quad question. anpanman is probably in sec 3 only and thus have no idea of differentiation.
For an quadratic equation of the form y= ax^2+bx+c...
min pt mean a>0, max pt mean a<0.
Thus all you have to do was equate a>0
ie (k-6)>0
k>6
Combine with -2<k<8 and you get 6<k<8.
Ahm97sic's answer is correct too. However he complicate the question unnecessarily.
Hi Mikethm,
I agree with your answer and comment.
Regards
ahm97sic
thanks to both ahm97sic and mikethm for the contributions :D
seems like we hav more and more maths experts coming to help in this forum :D