Find all the values of x where the tangent line is horizontal.
f(x) = 2x^3 + 42x^2 + 270x + 11

If I calculated correctly, the tangent line is horizontal where x ≈ -5.3 + 9.3i, and -5.3 - 9.3i

I'm somewhat concerned at having gotten complex numbers, and strongly recommend going through the steps to see if I missed anything.  I checked it myself and don't see any errors.

Step-by-step explanation:

You can do this by taking the derivative of the function and solving for zero:

f(x) = 2x³ + 32x² + 220x + 11

f'(x) = 6x² + 64x + 220

f'(x) = 2(3x² + 32x + 110)

We can't factor that further, so let's do it the long way, starting by letting f'(x) equal zero:

0 = 2(3x² + 32x + 110)

0 = 3x² + 32x + 110

0 = 9x² + 96x + 990

0 = 9x² + 96x + 256 + 734

0 = (3x + 16)² + 734

(3x + 16)² = -734

3x + 16 = ± i√734

3x = -16 ± i√734

x = (-16 ± i√734) / 3

x ≈ (-16 + 27.9i) / 3, and  (16 - 27.9i) / 3

x ≈ -5.3 + 9.3i, and -5.3 - 9.3i

I'm always wary when I end up with complex numbers.  I'd suggest double checking everything here, but I'm fairly certain I did everything correctly.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116