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This seems pretty simple and straightforward, but I can't seem to figure it out.

 

Give an example of a quadratic function that has zeroes at \(x=2\) and \(x=4\), and that takes the value \(6\) when \(x=3\).

Enter your answer in the expanded form "ax^2 + bx + c", where a,b,c are replaced by appropriate numbers.

 

Thank you very much for your help!

 Jun 14, 2020
 #1
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Zeroes are at x = 2 and x = 4 \(\Longleftrightarrow\) \(f(x) = k(x - 2)(x - 4)\)

 

In this case, f(3) needs to be equal to 6.

 

\(f(3) = k(3 - 2)(3 - 4) = -k\)

 

So -k = 6, and k must be -6.

 

\(f(x) = -6(x - 2)(x - 4)\)

 

You can finish it from here.

 Jun 14, 2020
 #2
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Got it! I appreciate the help.

 Jun 14, 2020

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