Question

The tangent to the curve $y = x^2$ at point (1, 1) is:

• A. $y = x$
• B. $y = 2x$
• C. $y = 2x - 1$
• D. $y = 2x -$

Answer 1

Answer: (C)

C) $y = 2x - 1$

Answer 2

To find the equation of the tangent line to the curve $y = x^2$ at the point (1, 1), we follow these steps:

1. Find the derivative of the function $y = x^2$. The derivative, denoted as $\frac{dy}{dx}$, gives the slope of the tangent line at any point on the curve.

   $$\frac{dy}{dx} = 2x$$

2. Evaluate the derivative at the given point (1, 1) to find the slope of the tangent line at that specific point. Substitute $x = 1$ into the derivative:

   $$m = 2(1) = 2$$

   Thus, the slope of the tangent line at (1, 1) is 2.

3. Use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:

   $$y - y_1 = m(x - x_1)$$

   where $(x_1, y_1)$ is the given point and $m$ is the slope. Substituting $(x_1, y_1) = (1, 1)$ and $m = 2$, we get:

   $$y - 1 = 2(x - 1)$$

4. Simplify the equation to obtain the equation of the tangent line in slope-intercept form:

   $$y - 1 = 2x - 2$$ $$y = 2x - 1$$

Therefore, the equation of the tangent line to the curve $y = x^2$ at the point (1, 1) is $y = 2x - 1$.