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Find the equation of the normal to the curve y=x²+4x-3 at point(1,2)

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Answer to a math question Find the equation of the normal to the curve y=x²+4x-3 at point(1,2)

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Maude

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First, find the derivative of the curve to get the slope of the tangent line. y = x² + 4x - 3 y' = 2x + 4 Evaluate the derivative at the point (1, 2) to find the slope of the tangent line at that point. y'(1) = 2(1) + 4 = 2 + 4 = 6 So, the slope of the tangent line at (1, 2) is 6. The slope of the normal line will be the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -1/6. Now that you have the slope of the normal line and the point (1, 2), you can use the point-slope form of a line to find the equation of the normal line: y - y₁ = m(x - x₁) Where: (x₁, y₁) = (1, 2) m = -1/6 (slope of the normal line) Now, plug in the values: y - 2 = (-1/6)(x - 1) Multiply both sides by 6 to eliminate the fraction: 6(y - 2) = -1(x - 1) 6y - 12 = -x + 1 Now, rearrange the equation to get it in standard form: x + 6y = 13

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