Finding the equation of the tangent I was working on the following question:

The curve $C$ has equation $y=(x+3)^2$ and the point $A$, with $x$ coordinate $-5$, and lies on $C$. 
* $a)$ Find the equation of the tangent line to $C$ at $A$, giving your answer in the form $y= mx+c$ 
* $b)$ Another point $B$ also lies on $C$. The tangents to $C$ at $A$ and $B$ are parallel. Find the $x$ coordinate of $B$. 

$\frac{d}{dx}(x+3)² = 2x+6 \text{ for } x = -5; 2x+6 = -4 = m$
I'm a bit lost :/
 A: For part $a$, you're halfway there. You've shown that $m = -4$ so $y = -4x +c$. To find $c$, we need to know a point on the line. All that we know about the line is that it is tangent to the point $A$ on the curve $C$. In particular, the point $A$ lies on the tangent line so we can use its coordinates to determine the value of $c$. So what are the coordinates of $A$? We are told the $x$-coordinate is $-5$, but what about the $y$-coordinate? Well $A$ does not only lie on the tangent line, but also on the curve $C$ which is defined by $y = (x + 3)^2$. Therefore $A$ has $y$-coordinate $y = (-5+3)^2 = (-2)^2 = 4$. So $C$ has coordinates $(-5, 4)$. You can now use this point to determine $c$ (by using the fact that $A$ is on the tangent line so its coordinates satisfy the equation of the tangent line).
For part $b$, use the fact that parallel lines have the same slope and that the slope of the tangent at $B = (x_0, y_0)$ to $C$ is given by $\frac{dy}{dx}(x_0)$. Furthermore, ignore the use of the word 'another' in the question, I think it is misleading.
A: For part (a):
As you've stated above, $\frac{dy}{dx}=2(x+3)=2x+6$ (by the chain rule). At the point with $x=-5, \frac{dy}{dx}=2(-5)+6=-4$ as you've said too. So the line you're looking for does indeed have $m=-4$.
What you need to do now is work out what $c$ should be. You know that the line $y=mx+c$ passes through $A$, and you can work out $A$'s $y$-coordinate, since it lies on $C$. Any point on the line satisfies $c=y-mx$ so you can work out $c$ in this way.
For part (b):
There is no different point $B$ with tangent to $C$ at $B$ parallel to the tangent to $C$ at $A$:
To say two lines are parallel is equivalent to saying that they have the same slope. So the tangent to $C$ at $B$ has equation $y=mx+d$ (where $d$ is as in (a)). This means that $\frac{dy}{dx}$ is the same at $A$ and $B$. So you're looking for another point on C with $\frac{dy}{dx}=-4$, i.e. with $2x+6=-4$. It's $x$-coordinate must also be $-5$. So $y=4$.
This gives the same point $A$ again.
A: The slope of the tangent line, at $A$ is equal to $y\prime(-5)$. Then you can just plug that back into the equation $y-y_1=m(x-x_1)$ and solve for $y$.
$$\frac{d}{dx}(x+3)^2 = 2(x+3) = 2x + 6 \implies y\prime(-5) = -10+6 = -4$$
$$y-y(-5) =y\prime(-5)(x+5)$$
$$y-4 = -4(x+5) \implies y-4=-4x-20$$
$$\boxed{y=-4x-16}$$
For part b, you have to find another point where the slope of the graph is $-4$, and therefore where $y\prime(x) = -4$
$$y\prime(x) = -4 \implies 2x+6 = -4 \implies 2x = -10 \implies x=-5$$
Therefore the only $x$ where the tangent has a slope of $-4$ is at $x=-5$. This means that in order to have a line tangent to $C$, and parallel to $A$, it must equal to $A$. This is why the word "another" is misleading, since there isn't a tangent line parallel to $A$ with different different $x$ and $y$ values where it is tangent to $C$.
Therefore, $B=A= (-5, 2)$ so the $x$ coordinate of the point is $-5$.
