1. ## Max or Min

The height of an object, h(t), is determined by the formula
h(t) = –16t2 + 256t, where t is time, in seconds. Will the object reach a maximum or a minimum? Explain or show your reasoning.

I have trouble with questions like this when the value of the variable is not given.

How can I find the answer without having to graph the given function?

2. Originally Posted by magentarita
The height of an object, h(t), is determined by the formula
h(t) = –16t2 + 256t, where t is time, in seconds. Will the object reach a maximum or a minimum? Explain or show your reasoning.

I have trouble with questions like this when the value of the variable is not given.

How can I find the answer without having to graph the given function?
You should recognise the equation as a parabola. And you should know how to find the turning point of a parabola.

3. ## Are you...

Originally Posted by mr fantastic
You should recognise the equation as a parabola. And you should know how to find the turning point of a parabola.
Are you saying that I should graph the given function?

4. No just take the derivative and use the first derivative test to see when it is a max or min or you could complete the square too. Why the large red font?

5. Originally Posted by magentarita
Are you saying that I should graph the given function?
No. But obviously if you're going to draw a graph their are features you try to calculate, including coordinates of the turning point. You don't have to draw the graph to get the coordinates.

Using symmetry, the t-coordinate of the turning point lies half-way between the t-intercepts. Do you know how to find it. Once you have the t-coordinate of the turnng point, how to get the h-coordinate shuld be clear. And it's a negative parabola (coefficient of squared term is negative). So the turning point is a maximum.

6. Originally Posted by magentarita
The height of an object, h(t), is determined by the formula
h(t) = –16t2 + 256t, where t is time, in seconds. Will the object reach a maximum or a minimum? Explain or show your reasoning.

I have trouble with questions like this when the value of the variable is not given.

How can I find the answer without having to graph the given function?
You are not asked to find the maximum or minimum, but just to identify if the given function has a max or min.

You should recognise that h(t) is a parabola opening downwards, and hence this has a maximum. Also you should know that the maximum occurs midway between the roots, which in this case are obvious.

RonL

7. h(t) = –16t2 + 256t

First derivative: -32t+256
Second derivative: -32 <----this is negative, graph is a parabola, concave DOWN

8. Originally Posted by lewisb13
h(t) = –16t2 + 256t

First derivative: -32t+256
Second derivative: -32 <----this is negative, graph is a parabola, concave DOWN
This is a pre-calculus forum, so methods relying on diffrentiation are not helpfull or in this case needed.

RonL

9. ## derivatives...

Originally Posted by 11rdc11
No just take the derivative and use the first derivative test to see when it is a max or min or you could complete the square too. Why the large red font?
We are not doing derivatives in our precalculus class.

I like to type using big words.

Originally Posted by mr fantastic
No. But obviously if you're going to draw a graph their are features you try to calculate, including coordinates of the turning point. You don't have to draw the graph to get the coordinates.

Using symmetry, the t-coordinate of the turning point lies half-way between the t-intercepts. Do you know how to find it. Once you have the t-coordinate of the turnng point, how to get the h-coordinate shuld be clear. And it's a negative parabola (coefficient of squared term is negative). So the turning point is a maximum.
How about if I graph the function y = = –16t^2 + 256t?

We are not doing derivatives in our precalculus class.

11. ## But....

Originally Posted by CaptainBlack
You are not asked to find the maximum or minimum, but just to identify if the given function has a max or min.

You should recognise that h(t) is a parabola opening downwards, and hence this has a maximum. Also you should know that the maximum occurs midway between the roots, which in this case are obvious.

RonL

How about if I graph the function y = = –16t^2 + 256t?

We are not doing derivatives in our precalculus class.

12. ## Look.....

Originally Posted by lewisb13
h(t) = –16t2 + 256t

First derivative: -32t+256
Second derivative: -32 <----this is negative, graph is a parabola, concave DOWN
How about if I graph the function y = = –16t^2 + 256t?

We are not doing derivatives in our precalculus class.

13. ## ok

Originally Posted by mr fantastic
No. But obviously if you're going to draw a graph their are features you try to calculate, including coordinates of the turning point. You don't have to draw the graph to get the coordinates.

Using symmetry, the t-coordinate of the turning point lies half-way between the t-intercepts. Do you know how to find it. Once you have the t-coordinate of the turnng point, how to get the h-coordinate shuld be clear. And it's a negative parabola (coefficient of squared term is negative). So the turning point is a maximum.
A negative parabola opens DOWNWARD.

A positive parabola opens UPWARD.

The downward parabola = minimum point...Is this true?

The upward parabola = maximum point...Is this true?

Thanks

14. ## ok

Originally Posted by CaptainBlack
You are not asked to find the maximum or minimum, but just to identify if the given function has a max or min.

You should recognise that h(t) is a parabola opening downwards, and hence this has a maximum. Also you should know that the maximum occurs midway between the roots, which in this case are obvious.

RonL
I fully understand.

Thanks

15. Originally Posted by magentarita
We are not doing derivatives in our precalculus class.

I like to type using big words.

O ok well in that case you can find the vertex like I said earlier by completing the square. Also remember a negative parabola opens down.