#8. find each point at which the tangent line to the curve is parallel to the line .

How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).

- Oct 15th 2009, 07:32 AMhazecrazeFinding points where a tangent lines is parallel to another line
**#8. find each point at which the tangent line to the curve is parallel to the line .**

How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5). - Oct 15th 2009, 07:42 AMstapel
Solve the second equation for "y=". Then read the slope value off the equation. (It'll be the value multiplied on the "x".)

Differentiate the first equation with respect to x. Set equal to the given slope value, and solve for the corresponding x-value(s).

Plug these x-values into the first equation, and solve for the corresponding y-value(s), and thus the point(s) in question.

(I get the same answer as you've listed.) (Wink) - Oct 15th 2009, 09:04 AMhazecraze
At first I confused my self by thinking it was and got (1,3), but I see that it was . Also, after differentiating and setting equal to -2 and then simplifying, I got . With synthetic division and a root, I got -(x-1)=0, which was the value of (1,7), but for the other part: , I factored out an to get , so wouldn't there also be a point 0, in addition to the -1, thus giving

*(0,1)*and then the (1,7) and (-1,-5)? - Oct 15th 2009, 09:40 AMramiee2010

on substituting in equation of curve we get

at (1, 7) and (-1, -5) ,the tangent line to the curve is parallel to the line http://img387.imageshack.us/img387/3531/graph.jpg - Oct 15th 2009, 10:31 AMhazecraze
Oh, well that was much more simplistic than what I did.