Question

What is the equation of the tangent to the curve y=x² at the point (1,1) ?

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

Answer 1

Answer: (B)

y = 2x - 1

Answer 2

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

1. Find the derivative of the curve's equation. The given curve is $y = x^2$. Differentiate $y$ with respect to $x$ to find the slope function: $\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x$

2. Calculate the slope of the tangent at the given point. The given point is (1,1). Substitute $x=1$ into the derivative to find the slope ($m$) of the tangent at this point: $m = \left.\frac{dy}{dx}\right|_{x=1} = 2(1) = 2$ So, the slope of the tangent line is 2.

3. Use the point-slope form of a linear equation. The point-slope form of a line is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope. Here, $(x_1, y_1) = (1,1)$ and $m=2$. Substitute these values into the formula: $y - 1 = 2(x - 1)$

4. Simplify the equation. $y - 1 = 2x - 2$ Add 1 to both sides to solve for $y$: $y = 2x - 2 + 1$ $y = 2x - 1$

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