
solving equation
http://mapleta7.eps.surrey.ac.uk:808...ejlhffhhaf.gif is the point on the curve http://mapleta7.eps.surrey.ac.uk:808...egmfpfgeod.gif where http://mapleta7.eps.surrey.ac.uk:808...hihknlolgd.gif.
The tangent to the curve at http://mapleta7.eps.surrey.ac.uk:808...ejlhffhhaf.gif has equation http://mapleta7.eps.surrey.ac.uk:808...nlkjhgcngp.gif.
What is the value of http://mapleta7.eps.surrey.ac.uk:808...bdhhmohaed.gif?
am not sure how to go about this question,
but so far i have done
$\displaystyle \frac{dy}{dx} = 2x  1 $
$\displaystyle y = 6x5 $
where do I go from here?
equation both equations?
I get $\displaystyle 2x2 = 6x5 $
I dont know what to do? Any help appreciated

Re: solving equation
Can I suggest that you try to keep clear in your mind that those two y's are not the same function.
You have two functions here, so let me given them separate names, f and l (for "line").
$\displaystyle f(x) = x^2  x + 5$, and $\displaystyle l(x) = 6x  5$.
Thus I'd translate the problem like this:
"Let the point $\displaystyle P = (k, f(k))$. Suppose the tangent line to $\displaystyle f$ at $\displaystyle P$ (i.e. at $\displaystyle x = k$) is $\displaystyle l$. Find $\displaystyle k$."
When I taught Calculus I, I think the phrase I repeated more than any other was "the derivative is the slope of the tangent line." When you take Calculus I, you should make sure that you deeply, completely, and unforgettably understand it. It's a central, vital concept.