Question

Determine all the points on the curve y = 2x3 + 3x2 - 18x + 3 where the slope of the tangent line is -6.

Answer 1

Consider the function $y=2x^3+3x^2-18x+3$

Differentiate with respect to $x$ as,

$\frac{dy}{dx}=\frac{d}{dx}(2x^3+3x^2-18x+3)$

$=2(3x^2)+3(2x)-18(1)+0$

$=6x^2+6x-18$

Set $\frac{dy}{dx}=-6$ (which is the slope of the tangent line) and solve for $x$ as,

$6x^2+6x-18=-6$

$x^2+x-3=-1$

$x^2+x-2=0$

$x^2+2x-x-2=0$

$x(x+2)-1(x+2)=0$

$(x-1)(x+2)=0$

$x=-2,1$

At $x=-2,y=35$ and at $x=1,y=-10$

Therefore, the points on the curve are: $(-2,35),(1,-10)$.