Thread: 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

If a line has equation
the parametric form is

So, how would I find p and q?

Originally Posted by r2d2

So, how would I find p and q?

is the slope of the line.

so P=-2 and q=-1?

So then the parametric equation should be,

<-2t-1,t-1>?

Originally Posted by r2d2

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

No indeed.

so