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.
Best Answer
Except for a negative sign, you are fine.
A negative sign is omitted when substituting (x,y) with (0,0)
The rest is correct.