Thread: Parallel and Tangent Lines, finding an equation

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

I've found so far that the slope of sqroot(x+1)=1/(sqroot(x+1) + sqroot(x+1))

as well as the equation of the parallel line to be y=x/6 + 2/3.

How do I solve for the other equation now?

Originally Posted by Jeavus

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

I've found so far that the slope of sqroot(x+1)=1/(sqroot(x+1) + sqroot(x+1))

as well as the equation of the parallel line to be y=x/6 + 2/3.

How do I solve for the other equation now?

you have an awkward way of writing the derivative. notice that

anyway, remember that the derivative gives the slope. since the slope of the line is 1/6, you want the tangent line passing through the point x such that the derivative is 1/6, that is:

solve for x, and you can find the coordinates of the point. then use that point in the point-slope form to find the equation of the line.

remember the point slope form says the equation of a line is given by:

where is the slope of the line and is a point the line passes through

Hello, Jeavus!

Your derivative is incorrect . . .

Find an equation of the straight line that is tangent to the graph of
and parallel to

The line is: . . . . Hence, its slope is

The derivative of the function is: .

So we have: .


Can you finish it now?

Originally Posted by Soroban

Hello, Jeavus!

Your derivative is incorrect . . .

not incorrect, just in an ugly format. it seems he used the limit definition to compute the derivative and didn't simplify afterwards