equation of tangent #2

**skeske1234** · August 5th 2009, 12:17 PM

Find the equation of the straight line that is tangent to the graph of f(x)=(x+1)^0.5 and parallel to x-6y+4=0

my attempt:
y=x^0.5+1^0.5
y'=0.5x^-0.5

M_t=0.5/sqrt(x)

not sure if I am doing it right and where to head from here..
please state the steps needed to finish this question. Thank you

**Plato** · August 5th 2009, 12:21 PM

You need to learn how to do derivatives.
$y = \sqrt {x + 1} \, \Rightarrow \,y' = \frac{1}<br /> {{2\sqrt {x + 1} }}$

**skeske1234** · August 5th 2009, 12:51 PM

Originally Posted by Plato

You need to learn how to do derivatives.
$y = \sqrt {x + 1} \, \Rightarrow \,y' = \frac{1}<br /> {{2\sqrt {x + 1} }}$

so with this derivative, how do i find the equation parallel to x-6y+4=0 now?

so far i have:

y=1/6x +b

**Plato** · August 5th 2009, 01:13 PM

Originally Posted by skeske1234

so with this derivative, how do i find the equation parallel to x-6y+4=0 now?

The slope of that line is $\frac{1}{6}$
So you need to find the value of x giving $\frac{1}{6}$ .
Solve $\frac{1}{{2\sqrt {x + 1} }} = \frac{1}{6}\, \Rightarrow \,x = 8$ .
Thus the point of tangency is $(8,3)$ .
Now write the tangent line.

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