Calculate a cubic curve from two slopes and two points I have a question that is supposed to be solved using derivatives. The question asks to find the cubic polynomial function $f(x)$ where $y=13(x-1)+4$ is tangent to $f(x)$ at $(1,4)$ and $y=(x+1)+6$ is tangent to $f(x)$ at $(-1,6)$
How do I go about finding the function $f(x)$ using this information and the fact that this is a question specifically about derivatives?
 A: Let the cubic be represented by the general form: $y(x) = ax^3 + bx^2 + cx + d$, where $a,b,c,d$ represent the coefficients.
You're given two points on the curve: $(1,4)$ meaning $y(1) = 4$ and $(-1,6)$, meaning $y(-1) = 6$.
So you can set up two equations:
$a + b + c  + d = 4$ (eq. 1)
$-a + b - c + d = 6$ (eq. 2)
Now you're also told the slopes at these two points. The equation of the tangent line is not that important - what's important is its slope at that point. So the slope at $(1,4)$ is $13$ and the slope at $(-1,6)$ is $1$ because those are the respective slopes of the tangent lines at those points.
The slope of the tangent to the curve at a variable $x$ value is given by $y'(x)$ (that's the derivative).
Here $y'(x) = 3ax^2 + 2bx + c$.
And by what you're given, $y'(1) = 13$ and $y'(-1) = 1$.
Now you can make another two equations:
$3a + 2b + c = 13$ (eq. 3)
$3a - 2b + c = 1$ (eq. 4)
$4$ linear simultaneous equations in $4$ variables. Can you do the rest?
A: It should be y′(1)= -1/13 and y′(−1)= -1, analysed from y=13(x−1)+4 is tangent to f(x) at (1,4) and y=(x+1)+6 is tangent to f(x) at (−1,6)
