<s>[INST] 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 :/
 [/INST] 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 $-5$.
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