A. find an equation for the line perpendicular to the tangent to the curve yequals=x cubed minus 9 x plus 7x3−9x+7 at the point (33,77). b. what is the smallest slope on the curve? at what point on the curve does the curve have this slope? c. find equations for the tangents to the curve at the points where the slope of the curve is 1818.

A. y=x³-9x+7; y'=3x²-9. Slope at (3,7) [I have assumed that (33,77) is meant to be (3,7)] = 3×9-9=18.
The perpendicular to the tangent has a slope of -1/18 and the equation is (y-7)=-(x-3)/18.
y=-x/18+1/6+7=-x/18+43/6.
B. When is 3x²-9 smallest? When 3x²-9=0, so x=±√3 and y=3√3-9√3+7=7-6√3 and y=7+6√3.
Points are (√3,7-6√3), (-√3,7+6√3).
C. I assume the slope is 18. 3x²-9=18, 3x²=27, x²=9, x=±3, and y=7. The equations are y-7=18(x-3) and y-7=18(x+3), so y=18x-47 and y=18x+61. These lines are parallel to one another.