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I found the derivative , which is the slope
is 1/ 2 sqrt (x+1)
what do i do next?

get the derivative of sqrt (x-1)...that's your equation of the tangent

what is the slope of
x-6y + 4= 0
?

your derivative is wrong, by the way...

is it.... cuz if it is parallel, i can just plug in that slope to the equation..

how is it wrong...

maybe it's just your notation\[\frac d{dx}\sqrt{x+1}=\frac d{dx}(x+1)^{1/2}=\frac12(x+1)^{-1/2}\]

yea..it looks the same
i have to solve it by limits

umm..isnt the answer of the derivative = the slope?

and for the line, i got 1/6 as the slope..it looks wrong...i dont know

so do i find x using 1/2 sqrt (1+x) = 1/6 ?

yes, you need to know the point x for the point-slope form when we make our line

oh ok thanks!!
i got x= 8 :)

so now you use the calculus version of point-slope form:\[y-y_1=f'(x_1)(x-x_1)\]

don't really get it ..