Question

The curve y=(k-6)x^2-8x+k cuts the x-axis at two points and has a minimum point. Find the range of values of k.

Answer 1

Hello,

answer: -2 < k < 8

Explanation:

As there are 2 roots: Δ>0

As there is a mininum, k-6 <0 ==> k<6,

minimum :y'=0 ==> (k-6)*2x-8=0 ==> x=4/(k-6)

`\Delta=8^2-4*k*(k-6)\\=64-4k^2+24k\\=-4(k^2-6k+9)+36+64\\=100-4(k-3)^2\\=4(8-k)(k+2)\\\\\Delta\ is\ positive\ for\ -2 < k < 8`