# calculus

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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

• calculus -

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)

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