calculus
posted by Anonymous .
find the point on the graph of the given function at which the slope of the tangent line is the given slope
x^3+9x^2+33x+15 slope= 6

You should have properly written the function as
y = x^3 + 9x^2 + 33x + 15
or
f(x) = x^3 + 9x^2 + 33x + 15
then dy/dx = 3x^2 + 18x + 33
but you want that to be 6
3x^2 + 18x + 33 = 6
x^2 + 6x + 9 = 0
(x+3)^2 = 0
x+3 = 0
x = 3
then f(3) = 27 + 81  99 + 15 = 30
The point is (3,30)