Determine an equation of the line that is tangent to the graph of f(x) = sqrt (x+1) and parallel to x-6y + 4= 0

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Determine an equation of the line that is tangent to the graph of f(x) = sqrt (x+1) and parallel to x-6y + 4= 0

Mathematics
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I found the derivative , which is the slope is 1/ 2 sqrt (x+1) what do i do next?
you need to find when that the derivative is the same as the slope of the given line what is the slope of the given line?
get the derivative of sqrt (x-1)...that's your equation of the tangent

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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
we need when that is the same as the slope of the given line. m:\[\frac12(x+1)^{-1/2}=m\]oh you need the definition, eh? ok...
\[\frac{\sqrt{x+1+h}-\sqrt{x+1}}h\]\[=\frac{\sqrt{x+1+h}-\sqrt{x+1}}h\cdot\frac{\sqrt{x+1+h}+\sqrt{x+1}}{\sqrt{x+1+h}+\sqrt{x+1}}\]\[=\frac{x+1+h-x-1}{h\sqrt{x+1+h}+\sqrt{x+1}}=\frac{h}{h\sqrt{x+1+h}+\sqrt{x+1}}\]\[=\frac{1}{\sqrt{x+1+h}+\sqrt{x+1}}\]which in the limit is\[\frac1{2\sqrt{1+x}}\]so what have you got for m ?
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
yes, and we want to know when that slope is the same as the slope of the given line\[\frac1{\sqrt{x+1}}=\frac16\]yes, 1/6 is what I got
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 ..
do u find a point and then plug in to y2- y1 /x 2- x1 = m?? i got 6y - x + 3 = 0 which is different from the textbook ans

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