Question

Find the equation of the tangent to the curve \(y = x^2\) at the point (1,1).

• A. \(y = 2x - 1\)
• B. \(y = 2x + 1\)
• C. \(y = x - 1\)
• D. \(y = x + 1\)

Hint:

The derivative at a point gives the slope of the tangent line at that point.

Answer 1

The Correct Option is A

The slope of the tangent is given by the derivative of \(y = x^2\): \[ \frac{dy}{dx} = 2x \] At \(x = 1\), the slope is \(2 \cdot 1 = 2\).
Using the point-slope form \(y - y_1 = m(x - x_1)\) at point (1,1): \[ y - 1 = 2(x - 1) \implies y - 1 = 2x - 2 \implies y = 2x - 1 \] Thus, option (1) is correct.