# On the curve y = x^2 - 4x + 3, at x = -1 find the equation of the normal line.

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The equation of the normal line on the curve y = x^2 - 4x + 3 at the point where x = -1 has to be determined.

For a curve y = f(x), the slope of the line tangent to the curve at a point where x = c is given by f'(c). As the normal is perpendicular to the tangent, the slope of the normal is -1/f'(c).

y = x^2 - 4x + 3

y' = 2x - 4

The slope of the tangent at x = -1 is -2 - 4 = -6. The slope of the normal at this point is 1/6.

If x = -1, y = (-1)^2 + 4 + 3 = 1 + 4 + 3 = 8

This gives the slope of the normal as (y - 8)/(x + 1) = 1/6

=> 6y - 48 = x + 1

=> 6y - x - 49 = 0

**The equation of the normal line on the curve y = x^2 - 4x + 3 at the point where x = -1 is 6y - x - 49 = 0**