# What is the equation of the tangent line at the point x=-2 to the curve y=x^2-5x+6 ?

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The slope of the tangent to a curve at a point is the value that the first derivative takes.

y = x^2 - 5x + 6

y' = 2x - 5

For x = -2

y' = -2*2 - 5 = -9

Also for x = -2, y = (-2)^2 + 5*2 + 6 = 4 + 10 + 6 = 20

The equation of the line with slope -9 passing through (-2, 20) is:

(y - 20)/(x + 2) = -9

=> y - 20 = -9x - 18

=> y + 9x - 2 = 0

**The required equation of the tangent is y + 9x - 2 = 0**

Since we only know the x coordinate of the tangency point, we'll determine the y coordinate:

y = (-2)^2 - 5*(-2) + 6

y = 4 + 10 + 6

y = 20

The coordinates of the tangency point are: (-2 ; 20).

We know that the expression of the first derivative represents the tangent line to the given curve.

dy/dx =(x^2 - 5x + 6)'

dy/dx = 2x - 5

If x = -2 => dy/dx = 2*(-2) - 5

dy/dx = -9

The slope of the tangent line is m = -9.

The equation of the tangent line, whose slope is m = -9 and the point of tangency is (-2 ; 20), is:

y - 20 = -9*(x + 2)

y = -9*(x + 2) + 20

y = -9x - 18 + 20

**The equation of the requested tangent line is: y = -9x + 2.**