<s>[INST] Finding the points on the curve where the gradient of the tangent is equal to 2. So, my question is as following.
"Find all the points on the curve $y= 2x^3 + 3x^2 - 10x +3$ where the gradient of the tangent is 2."
I used $\frac{dy}{dx}$ in order to get the formula for the gradient function.
$\frac{dy}{dx} = 6x^2 + 6x - 10$ (or when divided by 2 it's $3x^2 + 3x - 5$)
How do I continue? When I set the derived equation equal to 2, I do not get the answer given in the book.
Any input would be appreciated!
 [/INST] Why have you divided by 2?
I would have thought that the answer would be something like:
$dy/dx = 6x^2 + 6x - 10$
Then set this equal to 2:
$6x^2 + 6x - 10=2$
$\implies 6x^2 + 6x - 12=0 $
$\implies 2x^2 + 2x - 4=0 $
$\implies (2x+4)(x-1)=0 $
$\implies x=-2$ or $x=1$
Then substitute these values of $x$ in to your given equation for $y$ to give the points:
$y(-2)=2(-2)^3+3(-2)^2-10(-2)+3=2(-8)+3(4)+20+3=-16+12+23=19$
and $y(1)=2(1)^3+3(1)^2-10(1)+3=2(1)+3(1)-10+3=2+3-10+3=-2$.
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