At which point the slope to tangent is zero for the curvey `y=x^2-6x+8` ?

At what points the slopes of the tangents to the curve y = (x^(3))/(6) - (3x^(2))/(2) + (11x)/(2) + 12 increase ?

The slope of the tangent to the curve y=6+x-x^2 at (2,4) is

The slope of the tangent to the curve y=6+x-x^(2) at (2,4) is

If the slope of the tangent line to the curve y= (6)/(x^(2) - 4x + 6 ) at some point on it is zero, then the equation of the tangent is-

Find the slope of the tangent to the curve: y = 3x^2 - 6x at the point on it, whose x-co-ordinate is 2.

The slope of the tangent line to the curve x+y=xy at the point (2,2) is

Find the angle of inclination of the slope of the tangent to the curve x^(2)+2y=8x-7 at the point x=5.

1.Find the slope of tangent to the curve y=3x^(2)-6 at the point on it whose x - coordinate is 2.

Find the equation of the tangent line to the curvey =x^2-2x+7 which is perpendicular to the line 2y-x+1=0

Find the slope of the tangent to the curve : y=x^(3)-2x+8 at the point (1, 7) .