# Find the equation of the other tangent given the point of parallel tangent.

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.

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 $-3$.

• I understand now. Thank you so much! – Amanda Aug 25 '13 at 19:02