Additional Mathematics 0606 / Equations, inequalities and graphs / 3.963.96: Graphical inequality with y=∣x∣+3y = |x| + 3y=∣x∣+30606/21/O/N/25 — Question 2 · 3 marksMark as done · Save for later · Show all solutionsThe diagram shows the graph of y=∣x∣+3y = |x| + 3y=∣x∣+3.Use a graphical method to solve the inequality ∣x∣+3⩾∣2−x∣|x| + 3 \geqslant |2 - x|∣x∣+3⩾∣2−x∣.[3]▸ Answer▸ Official mark scheme← 3.95: Area between y=12−x2y = 12 - x^{2}y=12−x2 and y=x4−4x2+8y = x^{4} - 4x^{2} + 8y=x4−4x2+83.98: Reflection of a graph and inverse/composition of functions →
