# Math Help - tangent to graph and parallel to

1. ## tangent to graph and parallel to

find an equation of the line that is tangent to the graph of

$f(x) = \frac{1}{\sqrt{x}}$

and parallel to the line of $x + 2y - 6 = 0$

I know that

$\frac {d}{dx} \left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x^{3/2}}$

and that the parallel line can be rewritten as

$y = -\frac{1}{2}x+3$

but not sure how you use this to get the equation of the line that is tangent to the graph.

2. Originally Posted by bigwave
find an equation of the line that is tangent to the graph of

$f(x) = \frac{1}{\sqrt{x}}$

and parallel to the line of $x + 2y - 6 = 0$

I know that

$\frac {d}{dx} \left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x^{3/2}}$

and that the parallel line can be rewritten as

$y = -\frac{1}{2}x+3$

but not sure how you use this to get the equation of the line that is tangent to the graph.
The gradient of the line is $-\frac{1}{2}$. Parallel lines have the same gradient.

So the gradient of the graph will have to be $-\frac{1}{2}$.

Therefore

$f'(x) = -\frac{1}{2}$.

Solve for $x$.

3. ok, just from observation $x = 1$ and so $y = 1$

and the slope is $-\frac{1}{2}$ but when I try to build an eq I get

$y=-\frac{1}{2}\left(x-1\right)+1$

which appears to be the tangent we want but a little foggy still how this works. if indeed that is the answer

4. Originally Posted by bigwave
ok, just from observation $x = 1$ and so $y = 1$

and the slope is $-\frac{1}{2}$ but when I try to build an eq I get

$y=-\frac{1}{2}\left(x-1\right)+1$

which appears to be the tangent we want but a little foggy still how this works. if indeed that is the answer
You have $m = -\frac{1}{2}$ and $(x, y) = (1, 1)$.

So building the equation for the tangent:

$y = mx + c$

$1 = -\frac{1}{2}(1) + c$

$1 = -\frac{1}{2} + c$

$c = \frac{3}{2}$.

So the equation of the tangent is

$y = -\frac{1}{2}x + \frac{3}{2}$

or

$y = -\frac{1}{2}(x - 3)$.