# What is the slope of the line tangent to the curve y = x^3 + 3x^2 + 4x + 1 at (1, 9)

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### 1 Answer

You should write the equation of the tangent line to a curve `y = f(x)` , at a given point `A(x_A,y_A), ` such that:

`y - f(x_A) = f'(x_A)(x - x_A)`

You need to differentiate the given function with respect to x such that:

`f'(x) = 3x^2 + 6x + 4`

You need to evaluate f'(x) at x = 1 such that:

`f'(1) = 3 + 6 + 4 = 13`

You should substitute 9 for `f(x_A), ` 13 for `f'(x_A)` and 1 for `x_A` in equation of tangent line such that:

`y - 9 = 13(x - 1) => y = 13x - 13 + 9 => y = 13x - 4`

**Hence, evaluating the equation of tangent line to the curve `y = x^3 + 3x^2 + 4x + 1` , at the point (1,9), yields `y = 13x - 4.` **

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