The diagram shows a sketch of \( y = 18 + 5x – 2x^2 \).
(a) Find the coordinates of the points \( A, B \) and \( C \).
(b) Differentiate \( 18 + 5x – 2x^2 \).
(c) Find the coordinates of the point on \( y = 18 + 5x – 2x^2 \) where the gradient is 17.
▶️ Answer/Explanation
(a) \( A = (-2, 0) \), \( B = (0, 18) \), \( C = (4.5, 0) \)
For \( A \) and \( C \), set \( y = 0 \) and solve \( -2x^2 + 5x + 18 = 0 \) to get x-intercepts. For \( B \), set \( x = 0 \) to get y-intercept.
(b) \( 5 – 4x \)
Differentiate each term: derivative of 18 is 0, \( 5x \) becomes 5, and \( -2x^2 \) becomes \( -4x \).
(c) \( (-3, -15) \)
Set derivative equal to 17: \( 5 – 4x = 17 \). Solve to get \( x = -3 \). Substitute back into original equation to find y-coordinate.
(a) Solve the equation tan x = 11.43 for \(0\leqslant x\leqslant 360^{\circ}\)
(b) Sketch the curve \(y=x^{3}-4x\)
(c) A curve has equation \(y=x^{3}+ax+b\)
The stationary points of the curve have coordinates (2, k) and (-2, 10 – k).
Work out the value of a, the value of b and the value of k.
▶️ Answer/Explanation
(a) 85.0°, 265.0°
tan⁻¹(11.43) ≈ 85.0°. Tangent is positive in 1st and 3rd quadrants, so second solution is 85° + 180° = 265°.
(b) Cubic curve sketch
Shape crosses x-axis at x=-2,0,2 (roots of x³-4x=0). Has local max at x≈-1.15 and local min at x≈1.15.
(c) a = -12, b = 5, k = -11
1) Find dy/dx = 3x² + a. At stationary points (x=±2), dy/dx=0 → 3(4)+a=0 → a=-12.
2) Substitute (2,k) and (-2,10-k) into original equation to get two equations, solve simultaneously to find b=5 and k=-11.