You need to remember the form of equation of the tangent line to the graph of a given function `f(x)` , at a given point `(x_p,y_p)` , such that:

`f(x) - y_p = f'(x_p)(x - x_p)`

Let `f(x) = y` , such that:

`y - y_p = f'(x_p)(x - x_p)`

Since the problem provides the coordinates of the point ` (-4,1)` , yields:

`y - 1 = f'(-4)(x + 4)`

You need to evaluate the derivative of the function and you need to calculate `f'(-4)` , such that:

`x^2-4y^2=12 => 4y^2 = x^2 - 12 => y^2 = (x/2)^2 - 3`

`y = +-sqrt((x/2)^2 - 3)`

`y = f(x) => f'(x) = (((x/2)^2 - 3)')/(2((x/2)^2 - 3))`

`f'(x)= x/(4((x/2)^2 - 3)) => f'(-4) = -4/(4*sqrt(4 - 3)) => f'(-4) = -1`

Replacing `-1` for `f'(-4)` in equation of tangent line yields:

`y - 1 = -(x + 4) => y = -x - 4 + 1=> y = -x - 3`

**Hence, evaluating the equation of the tangent line to hyperbola `x^2 - 4y^2 = 12` , at the point `(-4,1)` , yields **`y = -x - 3.`

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