Gradient of the graph y = a - k/x

[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:

$$\frac{dy}{dx}$$ = -$$\frac{k}{x^{2}}$$ = -$$\frac{k}{(\frac{k}{a})^{2}}$$ = - $$\frac{k}{\frac{k^{2}}{a^{2}}}$$ = -$$\frac{ka^{2}}{k^{2}}$$ = -$$\frac{a^{2}}{k}$$
when y = 0

 

[DeanBH]

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