Gradient

DeanBH

" 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. =(

thx for help in advance

nicksauce

I imagine you would want to calculate dy/dx at that point by plugging in x = k/a into your expression for dy/dx.

Tedjn

How would you normally calculate the value of the derivative dy/dx at a point x?

cristo

Why are you all calculating derivatives? This is in the precalculus forum

Tedjn

The OP used derivatives, and it looks like a calculus problem. Maybe it should be moved?

DeanBH

didn't think it was worthy of the higher forum

i don't know how to simplify it properly, that's the problem. =(

DeanBH

can someone tell me how to simplify the answer, this isn't homework or coursework!

just revising need help! thanks

ineedmunchies

ah i think it should be:

[tex]\frac{dy}{dx}[/tex] = -[tex]\frac{k}{x^{2}}[/tex] = -[tex]\frac{k}{(\frac{k}{a})^{2}}[/tex] = - [tex]\frac{k}{\frac{k^{2}}{a^{2}}}[/tex] = -[tex]\frac{ka^{2}}{k^{2}}[/tex] = -[tex]\frac{a^{2}}{k}[/tex]
when y = 0

DeanBH

thought so, cheers beef