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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$1,2$

Maude
4.7
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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