Thread: Parallel and Tangent Lines, finding an equation

1. 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?

2. 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 $\displaystyle \frac 1{\sqrt{x + 1} + \sqrt{x + 1}} = \frac 1{2\sqrt{x + 1}}$

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: $\displaystyle \frac 1{2 \sqrt{x + 1}} = \frac 16$

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: $\displaystyle y - y_1 = m(x - x_1)$

where $\displaystyle m$ is the slope of the line and $\displaystyle (x_1,y_1)$ is a point the line passes through

3. Hello, Jeavus!

Your derivative is incorrect . . .

Find an equation of the straight line that is tangent to the graph of $\displaystyle f(x)\:=\:\sqrt{x+1}$
and parallel to $\displaystyle x-6y+4\:=\:0$

The line is: .$\displaystyle y \:=\:\frac{1}{6}x + \frac{2}{3}$ . . . Hence, its slope is $\displaystyle \frac{1}{6}$

The derivative of the function is: .$\displaystyle f'(x)\;=\;\frac{1}{2}(x+1)^{-\frac{1}{2}} \;=\;\frac{1}{2\sqrt{x+1}}$

So we have: .$\displaystyle \frac{1}{2\sqrt{x+1}} \:=\:\frac{1}{6}$

Can you finish it now?

4. 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