Find the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.
Practice Questions
1 question
Q1
Find the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.
y = 3x - 2
y = 2x + 1
y = 2x + 2
y = x + 3
f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 3). The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.
Questions & Step-by-step Solutions
1 item
Q
Q: Find 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, 3). The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.
Steps: 7
Step 1: Identify the function. The function given is y = x^2 + 2x.
Step 2: Find the derivative of the function. The derivative f'(x) = 2x + 2.
Step 3: Calculate the slope of the tangent line at x = 1. Substitute x = 1 into the derivative: f'(1) = 2(1) + 2 = 4.
Step 4: Find the y-coordinate of the point on the curve at x = 1. Substitute x = 1 into the original function: y = (1)^2 + 2(1) = 1 + 2 = 3. So the point is (1, 3).
Step 5: Use the point-slope form of the equation of a line. The formula is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point. Here, m = 4 and (x1, y1) = (1, 3).
Step 6: Substitute the values into the point-slope formula: y - 3 = 4(x - 1).
Step 7: Simplify the equation to get the slope-intercept form. Distributing gives: y - 3 = 4x - 4. Adding 3 to both sides gives: y = 4x - 1.
