Question 168096
Hi here i answered questions 2) and 3) contact me if you need more help

2. Calculate the points of the tangent of the curve : y = ex(x2 – 8) where and in which the tangent line is horizontal

{{{y=exp(x^2-8)}}} ,{{{dy/dx=2x*exp(x^2-8)}}}

y'=0 if only if x=0

then the point is (0,0) (where the tangent of the curve is horizontal)

3. Write one of the two tangents to the curve : y = 1 + x³ which are parallel to this equation 12x - y = 1.

{{{12x-y=1}}}
then
{{{y = 12x-1}}}

{{{y = 1+x^3}}} then {{{dy/dx=3x^2}}}

then 12 (first term of 12x-1 and y'(x)= 3x^2 are the same}

{{{12 = 3x^2}}} then {{{x^2=4}}} then {{{x=2}}} or {{{x=-2}}}

taking x=2 then equation of the line is {{{y = (dy/dx)*(2)(x-2)+y(2)}}} where {{{y(x)=1+x^3}}}

then line is {{{y = 12(x-2)+9=12x-15}}}

Answer: {{{y = 12x-15}}}