Question

horizontal tangents

Answer 1

At what value(s) of x does the graph \[h(x)=\frac{ 1 }{ 2 }x^2 \sqrt{2x+1}\]have a horizontal tangent? Show your work and give a reason for your answer

Answer 2

I took the derivative of h(x) and set it equal to zero and solve it.

Answer 3

I got x=0 and x=-2/5 not sure I did this correctly

Answer 4

Mmm I'm coming up with the same values!

Answer 5

These are critical points.
These are where the slope function `derivative function` is zero.
zero slope ~ horizontal tangent to the curve.
good good good.

Answer 6

Well we could be correct. I did not bring my TI89 home to check this

Answer 7

heh

Answer 8

Wolfram is kinda handy for checking your work.
https://www.wolframalpha.com/input/?i=y%3D.5x%5E2sqrt%282x%2B1%29
If you scroll down, you can see the max/min's listed.

Answer 9

The x=-1/2 is not coming from a horizontal tangent, it's from some other business.
Don't worry about that one.

Answer 10

ok, I started to think I did something incorrect

Answer 11

I am not too well versed with wolframalpha. Sometimes I will use it but I do everything old school and by hand.

Answer 12

Thanks :)

Answer 13

Here a graph of the function.
https://www.desmos.com/calculator/u0p0kqlwlm
See how at x=-1/2 we could actually draw a vertical line tangent to the function?
It's some other weird type of critical point there :)

Old school? haha fair nuff ^^

Answer 14

yes worked on my BS in math the good old way, with no calculator