Find the tangent line to:

$$f(x) = \sqrt{x-1}$$ that passes through the origin $(0, 0)$.

$$f'(x) = \frac{1}{2\sqrt{x-1}}$$

The line will be tangent at $(a, b)$ so then:

$$f'(a) = \frac{1}{2\sqrt{a-1}}$$

Which is the slope.

$$y - \sqrt{a-1} = \frac{1}{2\sqrt{a-1}}(x - a)$$

Through $(0, 0)$

$$\sqrt{a-1} = \frac{-a}{2\sqrt{a-1}}$$

$$a-1 = -\frac{a}{2}$$

Am I heading the RIGHT direction? I dont need an answer, just advice.


Except for a negative sign, you are fine.

A negative sign is omitted when substituting (x,y) with (0,0)

The rest is correct.

  • $\begingroup$ isnt point slope form: $$y - y_1 = m(x - x_1)$$? $\endgroup$ – Lebes Feb 15 '15 at 10:57
  • $\begingroup$ It is correct however fix the next equation to: $$-\sqrt{a-1} = \frac{-a}{2\sqrt{a-1}}$$ $\endgroup$ – Arashium Feb 15 '15 at 10:58

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