0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $-3$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.