Slope of a Tangent without given x value

Question

At what point on the graph of $y=-3x^3+2x-1$ is the tangent parallel to $y=2x+10$?

Now do I solve this question algebraically or do I solve it graphically since there is no specific x value given to find the slope of the tangent using the IROC method.

Answer 1

Let the tangent be the line $y=2x+c$

When we try to find the point where the line meets the curve, we will get a repeated root if the line is a tangent.

So $2x+c=-3x^3+2x-1$ needs to have a repeated root.

Rearrange to: $3x^3+c+1=0$

For there to be a repeated root, this must be written in the form $(x-a)^2(3x+b)=0$

Expand: $(x^2-2ax+a^2)(3x+b)=0$

$3x^3+(b-6a)x^2+(a^2-2ab)x+a^2b=0$

For the two expressions to be equivalent, there are certain cnditions that have to be met (compare coefficients).

$b-6a=0$

$a^2-2ab=0$

$a^2b=c+1$

Answer 2

Since the tangent line is parallel to $y=2x+10$ then it's of the form $y=2x + d$ for some real number $d$. Now plug this into the equation and see when it obtains a double/triple real root.

$$2x + d = -3x^3 + 2x - 1 \implies -3x^3 = d+1 \implies x^3=\frac{d+1}{-3}$$

This equation has a real triple root only when RHS is $0$, as otherwise we get two complex roots (multiples of the third roots of unity), so therefore we get that $d=-1$ and they touch each other at $(0,-1)$

Answer 3

The slope of the tangent line to the curve at $(x_0,y_0)$ is $$\frac{dy}{dx}_{\left|x=x_0\right.}=-9x_0^2+2$$ If there is a tangent line parallel to $y=2x+10$ it must hold $$-9x_0^2+2=2$$ Cause the slopes must be equal, so $x_0=0$. So, the point we was looking for is $(x_0,y_0)=(0,-1).$