Find all values of $x$ at which the tangent line to the given curve has intercept $ y= 2$ Find all values of $x$ at which the tangent line to the given curve has intercept $y = 2$
I am confused about the $y$-intercept $2$
the function $$f(x) = \frac{(2x + 5)}{(x + 2)}$$
The derivative is $$f'(x)  = \frac{- 1}{(x^2 + 4x + 4)}$$
Now what should I do next with the given $y$ intercept?
Sorry for a easy question just have misconception on derivative's relationships with a tangent and what is the derivative at a point gives us? 
 A: equation of tangent of f(x):
$$\forall t \in R,   y_x(t) =f'(x).t+b$$ 
Here $b=2$ because it is y-intercept 2 (go through (y,t)=(2,0)):
$$y_x(0) = f'(x).0 + b = 2$$
Finally you have to solve (go through f(x)):
$$y_x(x)=f(x)$$
or again:
$$ f'(x).x+2= f(x)$$
more explicit, you have to solve:
$$2-\frac{x}{x^2 + 4x + 4}=\frac{2x + 5}{x + 2}$$
A: The general equation of a line passing through ($x_1,y_1$) and having slope m is$$y-y_1=m(x-x_1)$$
The general equation of a tangent to the curve at ($x_1,y_1$) is,
$$Y-y_1=- \frac{1}{x^2 + 4x + 4}(X-x_1)$$where $(X,Y)$ is any point on the tangent.
The tangent line passes through (0,2). Hence substituting $X=0$ and $Y=2$ into the above equation, we get
$$y_1-2=\frac{1}{4}x_1$$
Now the slope of the tangent is $\frac {1}{4}$. 
Hence
$$\frac {1}{x^2 + 4x + 4}=\frac {1}{4}$$
Solve for x to get the required values.
A: At $\left(x_1, \dfrac{2x_1+5}{x_1+2}\right)$, the slope of the curve $y=\dfrac{2x+5}{x+2}$ is $\dfrac{-1}{(x_1+2)^2}$. So the equation of a line passing through that point with that slope (in point-slope form) is $$y-\frac{2x_1+5}{x_1+2}=\frac{-1}{(x_1+2)^2}\left(x-x_1\right)$$ 
If this line has $y$-intercept $2$, then it passes through $(0,2)$. Use that, and then see what $x_1$ has to be. There is only one solution for $x_1$, and hence only one point meeting the required condition.
