1Number 1ExplanationIf the tangent line to the curve y=x3−3x2−9xy = x^3 - 3x^2 - 9xy=x3−3x2−9x at point (a,b)(a, b)(a,b) has a gradient of 151515, then the possible value of a+ba + ba+b is....000−2-2−2−4-4−4−6-6−6−8-8−8ExplanationSubstitute point (a,b)(a, b)(a,b) into (x,y)(x, y)(x,y) in the equation b=a3−3a2−9a...(1)b = a^3 - 3a^2 - 9a \quad ...(1)b=a3−3a2−9a...(1) Find the first derivative f(x)=3x3−3x2−9xf(x) = 3x^3 - 3x^2 - 9xf(x)=3x3−3x2−9xf′(x)=9x2−6x−9f'(x) = 9x^2 - 6x - 9f′(x)=9x2−6x−9 Remember the concept that the first derivative of a function equals the gradient of the line f′(a)=mf'(a) = mf′(a)=m 3a2−6a−9=153a^2 - 6a - 9 = 153a2−6a−9=153a2−6a−24=03a^2 - 6a - 24 = 03a2−6a−24=0a2−2a−8=0a^2 - 2a - 8 = 0a2−2a−8=0(a+2)(a−4)=0(a + 2)(a - 4) = 0(a+2)(a−4)=0a=−2∪a=4a = -2 \cup a = 4a=−2∪a=4 Determine the value of bbb from equation (1)(1)(1) and a+ba + ba+b b=a3−3a2−9ab = a^3 - 3a^2 - 9ab=a3−3a2−9a When a=−2a = -2a=−2 b=(−2)3−3(−2)2−9(−2)b = (-2)^3 - 3(-2)^2 - 9(-2)b=(−2)3−3(−2)2−9(−2)b=−8−12+18b = -8 - 12 + 18b=−8−12+18b=−2b = -2b=−2a+b=−2+(−2)=−4a + b = -2 + (-2) = -4a+b=−2+(−2)=−4 When a=4a = 4a=4 b=43−3(4)2−9(4)b = 4^3 - 3(4)^2 - 9(4)b=43−3(4)2−9(4)b=−20b = -20b=−20a+b=4+(−20)=−16a + b = 4 + (-20) = -16a+b=4+(−20)=−16 Following the available options, the answer is −4-4−4.
