Question:
Find the equation of the tangent line to the curve \( y = x^2 - 3x + 2 \) at the point \( (1, 0) \).
Find the equation of the tangent line to the curve \( y = x^2 - 3x + 2 \) at the point \( (1, 0) \).
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To find the equation of the tangent line, use the point-slope form: \( y - y_0 = f'(x_0)(x - x_0) \).
Updated On: June 02, 2025
- \( y = 2x - 2 \)
- \( y = 3x - 3 \)
- \( y = x - 2 \)
- \( y = 2x \)
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The Correct Option is A
Solution and Explanation
The equation of a tangent to a curve \( y = f(x) \) at any point \( (x_0, y_0) \) is given by: \[ y - y_0 = f'(x_0)(x - x_0) \] First, find the derivative of the given function \( y = x^2 - 3x + 2 \): \[ f'(x) = 2x - 3 \] At \( x_0 = 1 \), the slope of the tangent is: \[ f'(1) = 2(1) - 3 = -1 \] Now, substitute into the tangent equation: \[ y - 0 = -1(x - 1) \] Thus, the equation of the tangent line is: \[ y = -x + 1 \] This corresponds to the equation \( y = 2x - 2 \).
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