Advanced Functions Question

M

Mumtaaz

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Explain how can you use the zeros of a quadratic function to help you determine where the function has a positive rate of change and a negative rate of change.

I have no clue how to answer this...
 
You will need more information than stated in the problem. It would be good to know if a>0 or a<0 where a is the coefficient of the x^2 term. Does this help? That is there should be two answers, one for a>0 and another for a<0.

If you know the two x-intercepts where is the line of symmetry, ie the x-value for the max or min value?
 
Jomo said:
You will need more information than stated in the problem. It would be good to know if a>0 or a<0 where a is the coefficient of the x^2 term. Does this help? That is there should be two answers, one for a>0 and another for a<0.

If you know the two x-intercepts where is the line of symmetry, ie the x-value for the max or min value?
Click to expand...
There is no other information other than the question, that's why I don't really know how I can explain in detail...
 
As I said you can NOT answer this question with just one answer. Your answer WILL depend on the a value. Let's change the problem.

Knowing that a>0 explain how can you use the zeros of a quadratic function to help you determine where the function has a positive rate of change and a negative rate of change.
 
Jomo said:
As I said you can NOT answer this question with just one answer. Your answer WILL depend on the a value. Let's change the problem.

Knowing that a>0 explain how can you use the zeros of a quadratic function to help you determine where the function has a positive rate of change and a negative rate of change.
Click to expand...
I can use the formula for average rate of change right?
 
Mumtaaz said:
Explain how can you use the zeros of a quadratic function to help you determine where the function has a positive rate of change and a negative rate of change …
Click to expand...
Hi Mumtaaz. What can you say is important about the x-value lying halfway between the two zeros?

?
 
Please draw two quadratic functions--one concave up (a>0) and one concave down (a<0). On each graph can you please label where the graph has a positive rate of change and a negative rate of change? Does your answer depend on the value of a? If not then you can do the problem as stated. If yes, then I am correct and you can't do the problem as stated. Whatever you do, please do not think that your teacher must be correct or that I must be correct. You need to see this for yourself! Please post back with your graph. The only way for you to get through this problem is by looking at the graph.
 
Jomo said:
Please draw two quadratic functions--one concave up (a>0) and one concave down (a<0). On each graph can you please label where the graph has a positive rate of change and a negative rate of change? Does your answer depend on the value of a? If not then you can do the problem as stated. If yes, then I am correct and you can't do the problem as stated. Whatever you do, please do not think that your teacher must be correct or that I must be correct. You need to see this for yourself! Please post back with your graph. The only way for you to get through this problem is by looking at the graph.
Click to expand...
I definitely understand the graph way I guess, I understand that a < x < 0 and 0 < x < b for the first graph, I also understand the second one.
 

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Mumtaaz said:
Explain how can you use the zeros of a quadratic function to help you determine where the function has a positive rate of change and a negative rate of change.

I have no clue how to answer this...
Click to expand...
I think we can do this in a general setting. Given two zeros p and q we can say that the quadratic is [math]f(x) = a(x - p)(x - q) = ax^2 - (p + q)x + pq[/math]. The derivative is [math]f'(x) = 2ax - (p + q)[/math]. We can solve [math]f'(x) = 0[/math] and since this is a quadratic we can say that f(x) is increasing/decreasing on one side or the other. You can classify these based on terms of a. (In other words, solve the problem and break it into two cases, one where a > 0 and one where a < 0.)

Can you fill in the steps from here?

-Dan
 
Otis said:
Hi Mumtaaz. What can you say is important about the x-value lying halfway between the two zeros?

?
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I feel like, relative to my graph, the x- value, 0, is the changing point for the positive and negative intervals
 
topsquark said:
I think we can do this in a general setting. Given two zeros p and q we can say that the quadratic is [math]f(x) = a(x - p)(x - q) = ax^2 - (p + q)x + pq[/math]. The derivative is [math]f'(x) = 2ax - (p + q)[/math]. We can solve [math]f'(x) = 0[/math] and since this is a quadratic we can say that f(x) is increasing/decreasing on one side or the other. You can classify these based on terms of a. (In other words, solve the problem and break it into two cases, one where a > 0 and one where a < 0.)

Can you fill in the steps from here?

-Dan
Click to expand...
I understand f(x) = a(x - p)(x - q) = ax^2 - (p + q)x + pq, are you saying that I can use this to explain the problem, also I should specify both cases for a since it isn't in the question right?
 
Mumtaaz said:
I definitely understand the graph way I guess, I understand that a < x < 0 and 0 < x < b for the first graph, I also understand the second one.
Click to expand...
The a I was referring to, as I already stated, is the one in the general formula for a quadratic equation which is ax^2 + bx + c.

I am not sure what the x means in a < x < 0 and 0 < x < b. You can move your graph to the left or right and the rate of change will not be different.

I also asked you to please show in each of your graphs to mark where the function has a positive rate of change and a negative rate of change.
 
topsquark said:
I think we can do this in a general setting. Given two zeros p and q we can say that the quadratic is [math]f(x) = a(x - p)(x - q) = ax^2 - (p + q)x + pq[/math]. The derivative is [math]f'(x) = 2ax - (p + q)[/math]. We can solve [math]f'(x) = 0[/math] and since this is a quadratic we can say that f(x) is increasing/decreasing on one side or the other. You can classify these based on terms of a. (In other words, solve the problem and break it into two cases, one where a > 0 and one where a < 0.)

Can you fill in the steps from here?

-Dan
Click to expand...
Can I please remind you of the distributive law? Even physicists accept that law.

Also this post is in a pre calculus forum
 
Jomo said:
The a I was referring to, as I already stated, is the one in the general formula for a quadratic equation which is ax^2 + bx + c.

I am not sure what the x means in a < x < 0 and 0 < x < b. You can move your graph to the left or right and the rate of change will not be different.

I also asked you to please show in each of your graphs to mark where the function has a positive rate of change and a negative rate of change.
Click to expand...
Oh, the intervals I gave indicated the positive and negative intervals for the first graph. Wouldn't any point on the left side of the graph give me a negative rate of change, and vice versa on the other side but instead of negative, positive?
 
No, the left side of both graphs do NOT always give a negative rate of change and to the right do NOT always give a positive rate of change.

Please look at each side of each graph one-by-one and decide whether the change is positive or negative. Please do not just assume that you know the answer. Get your hands a little dirty (but wash up afterwards because of the coronavirus) by looking at the graphs carefully.
 
Jomo said:
No, the left side of both graphs do NOT always give a negative rate of change and to the right do NOT always give a positive rate of change.

Please look at each side of each graph one-by-one and decide whether the change is positive or negative. Please do not just assume that you know the answer. Get your hands a little dirty (but wash up afterwards because of the coronavirus) by looking at the graphs carefully.
Click to expand...
No, I'm only speaking about the first graph, a > 0. For the second graph I think it maybe a negative rate of change on the left side and positive rate of change on the right.
 
Jomo said:
Can I please remind you of the distributive law? Even physicists accept that law.

Also this post is in a pre calculus forum
Click to expand...
I didn't note the Pre-Calc thing. Thanks.

And Physicists usually only accept Mathematical laws when they are convenient. :devilish: (Thanks for the catch!)

-Dan
 
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