# Parametric Tangent Problem

• Oct 28th 2010, 08:36 AM
r2d2
Parametric Tangent Problem
Te parametric form for the tangent line to the graph of y = 2x^(2)+2x-1 at x = -1 is

Ok, So I got the slope at x=-1, which is -2. Then I found a points along the line, (-1,-1).

So I have the slope of the tangent line, y= -2x-3.

How do I translate this into parametric form?

Thanks
• Oct 28th 2010, 09:01 AM
Plato
If a line has equation $\displaystyle y=\frac{p}{q}(x - x_0)+y_0,~q\not= 0$
the parametric form is $\displaystyle \left\{ \begin{gathered} x = qt + x_0 \hfill \\ y = pt + y_0 \hfill \\ \end{gathered} \right.$
• Oct 28th 2010, 10:00 AM
r2d2
So, how would I find p and q?
• Oct 28th 2010, 10:02 AM
Plato
Quote:

Originally Posted by r2d2
So, how would I find p and q?

$\displaystyle \dfrac{p}{q}$ is the slope of the line.
• Oct 28th 2010, 10:09 AM
r2d2
so P=-2 and q=-1?

So then the parametric equation should be,

<-2t-1,t-1>?
• Oct 28th 2010, 10:19 AM
Plato
Quote:

Originally Posted by r2d2
so P=-2 and q=-1?
So then the parametric equation should be,
<-2t-1,t-1>?

No indeed.

$\displaystyle \dfrac{p}{q}=\dfrac{-2}{1}$ so $\displaystyle <t-1,-2t-1>.$