Determine the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.

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Q: Determine the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.

Solution: f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 4). The tangent line is y - 4 = 4(x - 1) => y = 4x - 4 + 4 => y = 4x - 2.

Steps: 7

Step 1: Identify the function. The curve is given by the equation y = x^2 + 2x.
Step 2: Find the derivative of the function. The derivative f'(x) represents the slope of the tangent line. For y = x^2 + 2x, the derivative is f'(x) = 2x + 2.
Step 3: Evaluate the derivative at the point where x = 1. Substitute x = 1 into the derivative: f'(1) = 2(1) + 2 = 4. This means the slope of the tangent line at x = 1 is 4.
Step 4: Find the y-coordinate of the point on the curve when x = 1. Substitute x = 1 into the original equation: y = (1)^2 + 2(1) = 1 + 2 = 3. So the point on the curve is (1, 3).
Step 5: Use the point-slope form of the equation of a line to write the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope. Here, (x1, y1) = (1, 3) and m = 4.
Step 6: Substitute the values into the point-slope form: y - 3 = 4(x - 1).
Step 7: Simplify the equation to get it into slope-intercept form (y = mx + b). Distributing gives: y - 3 = 4x - 4. Adding 3 to both sides results in y = 4x - 1.