Find the equation of the tangent at the point (2, 0) to the curve y = x$^2$ - 2x

Analysis

To find the equation of a tangent to a curve at a specific point, we need two things:

1. The slope (gradient) of the tangent at that point.
2. The equation of a line using the point-slope form: $y - y_1 = m(x - x_1)$.

Concept

• Slope ($m$): The slope of the tangent to a curve $y = f(x)$ at any point is given by the derivative of the function, $\frac{dy}{dx}$, evaluated at that point.
• Point-Slope Form: Once we have the slope $m$ and the point $(x_1, y_1)$, we substitute them into the formula $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Find the derivative of the curve. Given the curve: $y = x^2 - 2x$ Differentiate with respect to $x$: $\frac{dy}{dx} = 2x - 2$

Step 2: Calculate the slope ($m$) at the point (2, 0). Substitute $x = 2$ into the derivative: $m = 2(2) - 2$ $m = 4 - 2$ $m = 2$

Step 3: Use the point-slope form to find the equation. We have $m = 2$ and the point $(x_1, y_1) = (2, 0)$. $y - 0 = 2(x - 2)$ $y = 2x - 4$

Conclusion

The equation of the tangent is $y = 2x - 4$. This matches option A.

Exam Tip

Always remember that the derivative $\frac{dy}{dx}$ represents the slope of the tangent at any point on the curve. If the question asks for the normal instead of the tangent, remember that the slope of the normal is the negative reciprocal of the tangent slope ($m_{normal} = -\frac{1}{m_{tangent}}$).