Find an equation of the tangent line to the graph of $f\left(x\right)=x^2+2x$ at the point where $x=1$.

• A. $y=2x+1$
• B. $y=4x-1$
• C. $y=3x+2$
• D. $y=x^2+2x$

Answer: (B)

Verified step by step guidance

1. Step 1: Recall that the equation of a tangent line to a curve at a given point is given by y = f'(x₀)(x - x₀) + f(x₀), where x₀ is the x-coordinate of the point of tangency.
2. Step 2: Compute the derivative of the function f(x) = x² + 2x to find f'(x). The derivative of x² is 2x, and the derivative of 2x is 2. Therefore, f'(x) = 2x + 2.
3. Step 3: Evaluate f'(x) at x = 1 to find the slope of the tangent line. Substitute x = 1 into f'(x): f'(1) = 2(1) + 2.
4. Step 4: Find the y-coordinate of the point of tangency by evaluating f(x) at x = 1. Substitute x = 1 into f(x): f(1) = (1)² + 2(1).
5. Step 5: Use the slope from Step 3 and the point (x₀, f(x₀)) from Step 4 to write the equation of the tangent line in point-slope form: y - f(x₀) = f'(x₀)(x - x₀). Simplify this equation to obtain the final equation of the tangent line.