1. ## Fun ol' Tangents

Hi,

Could I get some help with solving this. It's probably really simple but I'm not too sure how to approach it
______
'Find the equation of the tangent to the curve y=x√2x² + 7 at x=3'

Any help would be much appreciated.

Many thanks,
Moob

2. ## Re: Fun ol' Tangents

I can't read this. Is your function \displaystyle \begin{align*} y = x\,\sqrt{2}\,x^2 + 7 \end{align*} or \displaystyle \begin{align*} y = x\,\sqrt{2x^2} + 7 \end{align*} or even \displaystyle \begin{align*} y = x\,\sqrt{2x^2 + 7} \end{align*}?

3. ## Re: Fun ol' Tangents

Hi Prove It,

Sorry about the misunderstanding. It should be the last one.

Thanks a tonne!

4. ## Re: Fun ol' Tangents

Well the tangent line, being a line, will have equation \displaystyle \begin{align*} y = mx + c \end{align*}, so you need its gradient and its y-intercept. How will you find the gradient? How will you find the y intercept?

5. ## Re: Fun ol' Tangents

I've tried subbing 3 for x, this left me with
Attachment 28224

I am most likely way off in the method I'm using.

Thanks once again.

6. ## Re: Fun ol' Tangents

You need to find the derivative of the curve to get the gradient...

7. ## Re: Fun ol' Tangents

So would I replace m with this?

8. ## Re: Fun ol' Tangents

You would evaluated that at x = 3, and this gives you the slope, so use this along with the point (3,15) in the point-slope formula to determine the equation of the tangent line.

9. ## Re: Fun ol' Tangents

And no, the constant value is not 7. You will substitute x = 3, y = 15 and the value you find for m into y = mx + c and solve the equation for c.

10. ## Re: Fun ol' Tangents

Thank you both for all your help, I think I get it now!
If I have any problems I'll be sure to ask.

Thanks again!

11. ## Re: Fun ol' Tangents

Does this look about right to you?

Thanks a lot!

12. ## Re: Fun ol' Tangents

Yes, looks good to me.