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Find an equation of the tangent line to the parabola y = x² − 4x 8 at the point (1, 3).

User Winchestro
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Final answer:

To find the equation of the tangent line to the parabola y = x² − 4x + 8 at the point (1, 3), we need to find the slope of the tangent line. The slope of the tangent line is equal to the derivative of the function y = x² − 4x + 8 evaluated at x = 1. The equation of the tangent line is y = -2x + 5.

Step-by-step explanation:

To find the equation of the tangent line to the parabola y = x² − 4x + 8 at the point (1, 3), we need to find the slope of the tangent line. The slope of the tangent line is equal to the derivative of the function y = x² − 4x + 8 evaluated at x = 1. Taking the derivative of the function gives us y' = 2x - 4. Evaluating this at x = 1, we get y' = 2(1) - 4 = -2.

The slope of the tangent line is -2. Now, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line. Plugging in the values we have, we get y - 3 = -2(x - 1). Simplifying this equation gives us the equation of the tangent line as y = -2x + 5.

User Likk
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