Find the tangent

Question

"Find the equation of the two tangents that tangent to the function equation : "$y = \frac{x+3}{x+1}$" and are perpendicular (90degrees) to the graph function $2x-y=5$ (basicly simplified to $2x-5=y$)"

What I did is simplify $2x-y=5$ to $2x-5=y$ , I know that because its perpendicular than $y'=(x+3)+(x+1)$ and so $((2x+4)+b) (2x-5) = -1$

I do not know where to continue from here sadly. Any guidense would be great .

Tangent equations are supposed to be :

$y = -0.5x - 1.5$ and $y = -0.5x + 2.5 $

Answer 1

The function is

$$y=\frac{x+3}{x+1}=1+\frac2{x+1}\implies y'=-\frac2{(x+1)^2}$$

The above gives the slope of the tangent line at any point where the function is differentiable. Now, that slope must fulfill

$$\left(-\frac2{(x+1)^2}\right)\cdot2=-1\;\;\text{(why?)}$$

Well, now solve the above and find the $\;x\,-$ entry of the wanted point(s) ...

Answer 2

If you draw a tangent line at a point on the curve $(x+3)/(x+1)$, then its slope will be that of the derivative at that point. So, to begin with, let us take the derivative:

$$\frac{d}{dx}\frac{x+3}{x+1} = -\frac{2}{(x+1)^2}$$

If you cannot remember how to take this derivative, you could either use the quotient rule, or differentiate after rewriting the function as follows:

$$\frac{x+3}{x+1} = \frac{x+1+2}{x+1} = \frac{x+1}{x+1} + \frac{2}{x+1} = 1 + \frac{2}{x+1} = 1 + 2(x+1)^{-1} $$

The next constraint to satisfy is that your tangent line be perpendicular to the line $y = 2x - 5$. Recall that two lines are perpendicular if and only if the product of their slopes is $-1$. The given line has slope $2$, so you will need to find tangent lines with slope $-1/2$ as that yields $2(-1/2) = -1$.

So let us set the derivative equal to $-1/2$ and see for what $x$ values this occurs:

$$-\frac{2}{(x+1)^2} = -\frac{1}{2} \iff (x+1)^2 = 4 \iff x+1 = \pm 2$$

where the first iff is by cross-multiplication, and the second is by taking the square root of both sides of the equation. This yields solutions at $x=1$ and $x=-3$, each of which provides an answer since the original function is defined at both (it is defined everywhere except at $x= -1$).

So: All that is left is to figure out the equation of these two lines, noting that you have for each its slope (i.e., $-1/2$) and a point on it (i.e., the points on the original curve corresponding to $x = 1$ and $x = -3$). This is enough information to finish the problem.