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

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.