1 Answer | Add Yours
When line is tangent to the curve the gradient of tangent is equal to the gradient of line.
At tangent points;
`kx+6 = x^2+3x+2k`
`0 = x^2 +(3-k)x+(2k-6)` -------------(1)
At tangent points gradient is given by the first derivative of curve.
`dy/dx = 2x+3`
Since gradient of line = gradient of tangent;
`2x+3 = k rarr x= (k-3)/2`
`0 = ((k-3)/2)^2 +((3-k)(k-3))/2 +2k-6`
`0 = (k^2-6k+9)/4 - (k^2-6k+9)/2 +2k-6`
`0 = k^2-6k+9-(2k^2-12k+18)+8k-24`
So the required k values are k=11 and k=3
We’ve answered 318,989 questions. We can answer yours, too.Ask a question