" find, in terms of a and k, the gradient of the graph y = a - k/x at the point where it crosses x axis."

ok i worked out dy/dx = k/x^2 and x = k/a when y = o.

now what do i do. =(

 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 Recognitions: Homework Help Science Advisor I imagine you would want to calculate dy/dx at that point by plugging in x = k/a into your expression for dy/dx.
 How would you normally calculate the value of the derivative dy/dx at a point x?

Mentor

Why are you all calculating derivatives? This is in the precalculus forum
 The OP used derivatives, and it looks like a calculus problem. Maybe it should be moved?
 didn't think it was worthy of the higher forum i don't know how to simplify it properly, that's the problem. =(
 can someone tell me how to simplify the answer, this isn't homework or coursework! just revising need help! thanks
 ah i think it should be: $$\frac{dy}{dx}$$ = -$$\frac{k}{x^{2}}$$ = -$$\frac{k}{(\frac{k}{a})^{2}}$$ = - $$\frac{k}{\frac{k^{2}}{a^{2}}}$$ = -$$\frac{ka^{2}}{k^{2}}$$ = -$$\frac{a^{2}}{k}$$ when y = 0

 Quote by ineedmunchies ah i think it should be: $$\frac{dy}{dx}$$ = -$$\frac{k}{x^{2}}$$ = -$$\frac{k}{(\frac{k}{a})^{2}}$$ = - $$\frac{k}{\frac{k^{2}}{a^{2}}}$$ = -$$\frac{ka^{2}}{k^{2}}$$ = -$$\frac{a^{2}}{k}$$ when y = 0
thought so, cheers beef