Thread: Find Co-ords of parallel tangent to quad function

Looking for the co-ordinates of the tangent parallel to this when x=0.6

2x^3-6x^2+10x+9

so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)

how do i find the parallel tangent and then its co-ordinates.

You mean you're looking for the equation of the tangent of the curve?

A 2nd tangent parallel to one at (0.6,13.272)

Originally Posted by stophe73

Looking for the co-ordinates of the tangent parallel to this when x=0.6

2x^3-6x^2+10x+9

so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)

how do i find the parallel tangent and then its co-ordinates.

Originally Posted by stophe73

A 2nd tangent parallel to one at (0.6,13.272)

I don't get this either. You are looking for "the co-ordinates of the tangent parallel to this" y =
2x^3-6x^2+10x+9 at the point (0.6, 13.272)? We can certainly get the tangent line to this curve at x = 0.6. Is this what you are looking for? What do you mean by "2nd tangent?" This is badly worded.

-Dan

this function(graph is s shaped 2 knees? turns?) has a tangent to it at co ords 0.6,13.272 with gradient 4.96.
I'm looking for another tangent with a parallel gradient , but different co-ordinates.

I can only interpret this as meaning that you are looking for a value of x, other than .6 that has the same slope.

With , . When , that is


so now you are looking for another value of x that satisfies
That is the same as solving and that is especially easy to solve because you already know that x= .6 is a solution: divide by .

The question asks for the x,y on the same 'line' where the same gradient / slope occurs. i.e the co-ordinates where a parallel tangent exists.

So how would I carry out it out for the y, co-ordinate? ('original' y by calculation 13.272)

From the post of HallsofIvy, you should see that you will get another value of x where a tangent will have the same gradient as the tangent at x = 0.6.

Once you get the other x coordinate, you can find the other coordinate.

You mean with y=mx+c , how do i find c for the "2nd tangent" , sorry about the confusing grammar.

No, I mean you look for the other value of x which satisfies the equation given by HallsofIvy:



One solution is 0.6, as given by your question. There is another one. Find it, then use that x to get the y-coordinate.

I see...but how do i find the y co-ord and the intercept?

You do just like you did for the first one, you use the value of that new x and plug it in the original equation.

Then, you take the point and the gradient of the tangent to get directly the equation of the line.