# `f(x) = x^2 + 3, (-1, 4)` Find an equation of the tangent line to the graph of f at the given point.

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### 1 Answer

The given function is:-

f(x) = (x^2) + 3

differentiating both sides w.r.t 'x' we get

f'(x) = 2x + 0

Now, slope of the tangent at the point (-1,4) = f'(-1) = 2*(-1) = -2

Thus, equation of the tangent at the point (-1,4) and having slope = -2 is :-

y - 4 = (-2)*(x - (-1))

or, y - 4 = -2x - 2

or, y + 2x = 2 is the equation of the tangent to the given curve at (-1,4)