Given w(x) = -2x^2 + 16 + a, find "a" so absolute max of w(x) is 39.

nanluo

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If anyone could help with this question it would be great thanks!

Give a function w(x)=-2x2+16+a. What is the value of a if the absolute maximum value of the function is 39?

Thanks!
 
If anyone could help with this question it would be great thanks!

Give a function w(x)=-2x2+16+a. What is the value of a if the absolute maximum value of the function is 39?

Thanks!
What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

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If anyone could help with this question it would be great thanks!

Give a function w(x)=-2x2+16+a. What is the value of a if the absolute maximum value of the function is 39?

Thanks!

It would be particularly helpful to know whether you have learned any calculus, or know how to find the vertex of a parabola, or whatever else this might be intended to use. That's part of the reason we need to see your work (in addition to seeing whether you do that work correctly).
 
Your question has been moved from "News" (the first forum on the page) to an appropriate category. Also, your subject has been changed from "Graph theory" (which doesn't relate at all to this question) to something more relevant. For future reference, kindly please note that considerate posting generally results in better (and faster) responses.

Give a function w(x)=-2x2+16+a. What is the value of a if the absolute maximum value of the function is 39?
Complete the square to convert the quadratic equation into vertex form, w(x) = -2(x - h)^2 + k. Then note the shape of the parabola (because of the sign of the leading coefficient). Set the relevant portion of the completed square equal to the given value. Solve the resulting linear equation for the value of "a".

If you get stuck, please reply showing your thoughts and efforts so far. Thank you! ;)
 
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