Additional Mathematics 0606 / Calculus — Differentiation 1 / 3.543.54: Non-zero k for a tangent condition0606/22/O/N/23 — Question 2 · 5 marksMark as done · Save for later · Show all solutionsFind the non-zero value of kkk for which the line y=−2x−6−1ky = -2x - 6 - \dfrac{1}{k}y=−2x−6−k1 is a tangent to the curve y=kx(x+2)y = kx(x + 2)y=kx(x+2).[5]▸ Answer▸ Official mark scheme← 3.53: Normal determining constants p and q3.55: Normal to a cubic curve →
