How Do I Find the Gradient of y=\frac{5x-4}{x^2} at the X-Axis Crossing?

[footprints]

Find the gradient of the curve $$y=\frac{5x-4}{x^2}$$ at the point where the curve crosses the x-axis.
After I differentiating the equation, I got $$-\frac{5}{x^2} + \frac{8}{x^3}$$ (it might be wrong). Now what do I do?

 

[Nylex]

Well, dy/dx gives you the gradient at x. The curve crosses the x-axis when y = 0. You've worked out the derivative correctly though.

 

[footprints]

The curve crosses the x-axis when y = 0

I thought so too. But I couldn't get the right answer.

 

[Nylex]

I thought so too. But I couldn't get the right answer.

What answer did you get?

 

[footprints]

$$3\frac{1}{8}$$

 

[Nylex]

$$3\frac{1}{8}$$

Can you post your working? I got dy/dx = 7.8125. Also, what value did you get for the x intercept?

 

[footprints]

To get x I must substitue y=0 into $$y=\frac{5x-4}{x^2}$$ right?
Btw the answer my book gives me is the same as yours

 

[Nylex]

To get x I must substitue y=0 into $$y=\frac{5x-4}{x^2}$$ right?
Btw the answer my book gives me is the same as yours

Yes, then solve for x.

y = (5x - 4)/x^2

=> (5x - 4)/x^2 = 0

Need to solve for the numerator being equal to 0, so x = 4/5.

 

[footprints]

Thats great. Then I plugged in 4/5 into
$$-\frac{5}{x^2} + \frac{8}{x^3} \rightarrow -\frac{5}{\frac{4}{5}^2} + \frac{8}{\frac{4}{5}^3}$$
Then I solve and get $$3\frac{1}{8}$$

 

[footprints]

Oh now I know why I didn't get the answer. I calculated a part wrong. Sorry. Thank you for your time.

 

[Nylex]

No problem .