please help!

Question

please help!

anonymous · Answer

attachment

directrix · Answer

@Dodo1  Is the exponent of the e term in the problem (-6x) or (-8x)?

anonymous · Answer

-6x!

kainui · Answer

Ok, so tell me what you understand about critical points Dodo, and I will make sure you understand them completely so these types of problems are a breeze. =)

anonymous · Answer

Thank you! I understand that I have to do...
factorize then (x-1)=0.  use qudartic fomrula to solve out?

kainui · Answer

This is a calculus class, right?

kainui · Answer

The quadratic formula shouldn't be necessary for this problem as far as I can tell.

anonymous · Answer

Yes calculus 1. Oh I see.

kainui · Answer

|dw:1364445013848:dw| So a critical point is just a place where you can draw a tangent line that's completely horizontal, like this:

kainui · Answer

Critical points are at the tops of humps and bottoms of humps. They always have a slope of 0 there.

anonymous · Answer

i see! thanks for your drawing

kainui · Answer

So if you take the derivative and set it = to zero you can solve for the values which are critical points on the original graph.

kainui · Answer

So, suppose you want to find the critical point of y=x^2 (the normal parabola we all know) 

Take the derivative:

dy/dx=2x

Now you want to find where the slope is 0, so dy/dx=0

0=2x

Solve for x by dividing both sides by 2,

x=0

This is where the graph has its lowest point, right? Makes sense off of what we already know.

anonymous · Answer

ok!

anonymous · Answer

so i have to do derivative first then solve for x