Finding the points on the curve where the gradient of the tangent is equal to 2.

Question

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!

Answer 1

We know that $\frac{dy}{dx} = 6x^2 + 6x - 10$. We can factor out a $2$ from the polynomial, giving us $\frac{dy}{dx} = 2(3x^2 + 3x - 5)$; substituting the desired slope of $2$ gives us the quadratic $2(3x^2 + 3x - 5) = 2$. You can't just divide the polynomial by a constant factor when the other side is nonzero unless you also divide the other side by the same factor.

Answer 2

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 $y(1)=2(1)^3+3(1)^2-10(1)+3=2(1)+3(1)-10+3=2+3-10+3=-2$.