Gradient of the tangent (differentiation)

[IloveManUtd]

Find the gradient of the tangent to the curve y = (2x-1)/x[sup:1x8n67k8]2[/sup:1x8n67k8]+3 at the point where the curve cuts the x-axis.

I've found the dy/dx already but I'm not sure what I should do next.

 

[mmm4444bot]

Something is wrong somewhere, with this exercise.

The graph of y = (2x-1)/x2 + 3 has two x-intercepts, not one.

You were savvy enough to put grouping symbols around the numerator, so I have no expectation that the denominator is anything other than x^2.

Do you have additional (i.e., undisclosed) information (eg: a domain statement) that would reduce the scenario to one point on the x-axis ?

Also, why do you not show work that you've already done (or, at least, the result) ?

I'm thinking that by now you've been asked a few times to show your work. Doing so might significantly lessen the aggregate typing associated with each thread, and that would get you to where you need to be sooner.

< sigh >

Regardess, you need to find the x-coordinate at the point of tangency (i.e., an x-intercept, where it goes without saying -- yet, I'm saying it anyway -- y = 0).

Evaluate the derivative, using that value of x, to determine the slope of the curve at the point of tangency.

(That's the slope of the tangent line, too.)

That's what they're asking for, yes?

Cheers

Why do you still type x2 for x^2 ?