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Find an equation of the tangent line to the parabola y = x² - 4x + 6 at the point (1, 3).

User Jackyto
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Final answer:

To find the equation of the tangent line to the parabola y = x² - 4x + 6 at the point (1, 3), we need to find the slope of the tangent line by taking the derivative of the function at that point. The derivative is 2x - 4, so the slope at x = 1 is -2. Using the point-slope form of a line, we can write the equation of the tangent line as y = -2x + 5.

Step-by-step explanation:

To find the equation of the tangent line to a parabola at a given point, we need to find the slope of the tangent line. The slope of the tangent line to the parabola y = x² - 4x + 6 at the point (1, 3) is the derivative of the function at that point. Therefore, we need to find the derivative of y = x² - 4x + 6 and evaluate it at x=1. The derivative of y = x² - 4x + 6 is y' = 2x - 4, and when x = 1, the slope is 2(1) - 4 = -2.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point (1, 3) and m is the slope, we can substitute the values to find the equation of the tangent line. Plugging in the values, we get y - 3 = -2(x - 1).

Simplifying, we have y - 3 = -2x + 2, which can be rewritten as y = -2x + 5. Thus, the equation of the tangent line to the parabola y = x² - 4x + 6 at the point (1, 3) is y = -2x + 5.

User Andreas Siegel
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