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1. To determine the tangent line to the curve defined by the equation y = x^2 + 3x + 2 at the specific point where x equals 1, one must first evaluate the function at this point. By substituting x = 1 into the equation, we can calculate the corresponding y-coordinate. This results in y = (1)^2 + 3(1) + 2, which simplifies to y = 6. Therefore, the point of tangency on the curve is (1, 6).
2. Next, it is essential to find the slope of the tangent line at the point of interest. This requires computing the derivative of the function, which represents the slope of the curve at any given point. The derivative of the function y = x^2 + 3x + 2 is given by dy/dx = 2x + 3. By substituting x = 1 into this derivative, we find the slope at that point to be dy/dx = 2(1) + 3, resulting in a slope of 5.
3. With both the point of tangency and the slope established, we can now formulate the equation of the tangent line. Utilising the point-slope form of a linear equation, which is expressed as y – y1 = m(x – x1), where (x1, y1) is the point of tangency and m is the slope, we substitute our values: y – 6 = 5(x – 1). Simplifying this equation leads to y = 5x + 1. Thus, the equation of the tangent line to the curve at the specified point is y = 5x + 1.
