Find the equation of the other tangent to $y = 1 – 3x + 12x^2 – 8x^3$ which is parallel to the tangent at $(1,2)$.
I have already found the slope of the lines using the given point:
$$\begin{align}
y &= 1 – 3x + 12x^2 – 8x^3 \\
\implies y' &= -3 + 24x – 24x^2 \\
\implies y' &= -3 + 24(1) – 24(1)^2 \\
\implies y' &= -3
\end{align}$$
My question is how to find the point at which the other tangent lies.
Best Answer
You know the slope of the tangent for any point $(x,y)$ on the curve and that is given by $$ y' = -3 + 24x - 24x^2$$
You also know that the tangent that you are required to find has the slope of $-3$. So, let the point you are looking for be some $(x_1, y_1)$. The slope of tangent at this point is:
$$ \text{slope} = -3 + 24x_1 - 24x_1^2$$
This slope is equal to $-3$. So,
$$ -3 + 24x_1 - 24x_1^2 = -3 \implies x_1 = 0 \text{ or } x_1 = 1$$
Therefore the corresponding $y_1$ values are $ y_1 = 1 \text{ or } y_1 = 2 $ (By subbing $x = 0$ and $x = 1$ in the curve equation)
So, there are two points on the curve, the tangent at which has the slope of $-3$. In other words, the tangents at these two points are parallel to the tangent at the point $(1,2)$. And the two points are:
$$ P(0,1) \text{ and } Q(1,2) $$
Obviously $Q$ is the given point. So we are interested in $P$. Your answer is the line passing through $P$ and that has a slope of $-3$.