What is the equation of the tangent line to the curve y = x^2 + 2x at the point (1, 3)?

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Q: What is the equation of the tangent line to the curve y = x^2 + 2x at the point (1, 3)?

Solution: f'(x) = 2x + 2. At x = 1, f'(1) = 4. The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.

Steps: 6

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 of interest. We need to find the slope of the tangent line at x = 1. So, calculate f'(1) = 2(1) + 2 = 4.
Step 4: Use the point-slope form of the equation of a line. The point-slope form is y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope. Here, (x1, y1) = (1, 3) and m = 4.
Step 5: Substitute the values into the point-slope form. This gives us y - 3 = 4(x - 1).
Step 6: Simplify the equation to get the slope-intercept form. Distributing gives y - 3 = 4x - 4, and adding 3 to both sides gives y = 4x - 1.