# Thread: equation of tangent #2

1. ## equation of tangent #2

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

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

3. Originally Posted by Plato
You need to learn how to do derivatives.
$y = \sqrt {x + 1} \, \Rightarrow \,y' = \frac{1}
{{2\sqrt {x + 1} }}$
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

4. 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.