# Find the slope of the tangent to the graph of the function y = 3 + 2x-x ^ 2 at the point with the abscissa x = 1.

Let us give the equation of the tangent line in general form: y = y0 + f ‘(x0) * (x – x0), where y0 = f (x0), and f’ (x0) is the value of the derivative f ‘(x) at the point x0. By the condition of the task x0 = 1, then y0 = 3 + 2 * 1 – 1² = Since f (x) = 3 + 2 * x – x², then we have: f ‘(x) = (3 + 2 * x – x²) ‘= 0 + 2 * 1 – 2 * x = 2 – 2 * x. We calculate f ‘(x0) = 2 – 2 * 1 = 0. Therefore, the equation of the tangent line drawn to the graph of this function at a point with abscissa x0 = 1 has the form: y = 4 + 0 * (x – 1) or y = 4. As you know, the general equation of a straight line with a slope is: y = k * x + b, where k is the slope. This means that the desired slope is 0.

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