Question

how to find the turning point of T(x)=(x+5)^3+7

Please Include step by step instruction. I don't just want to cheat I actually want to know what I'am doing.

Answer 1

Turning point is when T'(x) = 0. So u need to differentiate T(x) - do u know how to do that?

Answer 2

Osrry but I'm not sure whatyou are talking about...

Answer 3

t(x) is just like f(x)

Answer 4

In order to find turning point, u need to know differentiation so as to calculate dT(x)/dx....

Answer 5

dave is talking about derivatives

Answer 6

But I don't think this person knows derivatives based on her other problems.

Answer 7

Yes, T(x) is just like f(x) but for the turning point u need to find dT(x)/dx=0

Answer 8

@ChristopherToni Oh...thanks...then I probably should wait for someone else to help because i dont know how to do w/o differentiation :( sorry @Memyselfmaria

Answer 9

its okay, sorry I'm a little slow on math....

Answer 10

is anyone going to help?

Answer 11

Turning points occur where your function has a maximum or a minimum (i.e. a point where the function goes from increasing to decreasing or vice versa). In calculus, that's where the derivative is zero but that's waaaaaay beyond the scope of what you're currently doing. In general, there usually isn't a way to find the exact values of the turning without calculus, but for some functions, we can compute them without calculus; for instance, we can compute the turning point for a quadratic equation (degree 2 polynomial) because the turning point is the same as the vertex. Lucky for us, though, \(y=(x+5)^3+7\) is easily obtained by translating \(y=x^3\) 5 units to the left and 7 units up. If you look at the graph of \(y=x^3\), you'll see that the function is always increasing and hence there's no turning point. Therefore, \(y=(x+5)^3+7\) will also be always increasing and thus has no turning points.

I hope this makes sense! :-)

Answer 12

what do you mean by translating? I have to like explain this with complete sentences and I'm lost on hhow to do that.

Answer 13

@Memyselfmaria can u solve this by using a graphic calculator like: http://www.desmos.com/calculator?

u can show the graph n identify the turning point from there.

Answer 14

What I mean is that if you start with \(\large y=x^3\), then \(\large (x+5)^3\) is a horizontal shift 5 units to the left and thus \(\large (x+5)^3+7\) has a horizontal shift 5 units to the left and a vertical shift of 7 units up:



Thus, if we determine the turning points of \(\large y=x^3\), then we had the turning points of \(\large y=(x+5)^3+7\) (via translation). However, since \(\large y=x^3\) has no turning points due to the fact that it's always increasing, then it follows that it's translation \(\large y=(x+5)^3+7\) also has no turning points (since it too is always increasing).

Does this clarify things? :-)