find all the values of x where the tangent line is horizontal.
f(X)=2x^3+36x^2+192x+5

Question

find all the values of x where the tangent line is horizontal. 
f(X)=2x^3+36x^2+192x+5

anonymous · Answer

You'll need to find the derivative of the function given because that'll give you the equation of slope right?

Then set that equation equal to 0 and solve for x!

anonymous · Answer

still lost

anonymous · Answer

Well, let's find the derivative first:
$$2x ^{3}+36x ^{2}+192x+5$$
take that derivative to get:
$$6x ^{2}+72x+192$$
does that make sense?

anonymous · Answer

Since the derivative is the equation of the slope, you can set the derivative equal to 0, which means the slope is 0 meaning that the tangent line is horizontal, yes?

anonymous · Answer

ok

anonymous · Answer

So what we have to do is find the x coordinates that the derivative crosses the x-axis
$$6^{2}+72x+192=0$$
then you factor out a 6:
$$6(x ^{2}+12x+32)$$
then factor again, setting it equal to 0
$$6(x+4)(x+8)=0$$

anonymous · Answer

this is how far i got

anonymous · Answer

Well, now you know that the slope of the original function is 0 at -4 and -8
which means that the tangent line is horizontal at -4 and -8