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anonymous
 4 years ago
Describe the transformation for : y= x²+6x5
anonymous
 4 years ago
Describe the transformation for : y= x²+6x5

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The parent function is the squaring function, \(y=x^2\) Look at all the ways \(y= x^2+6x5\) is different from the parent function. Putting it in vertex form will also help.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how do i do that ? :$

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Vertex form? Complete the square.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\huge Range = {\epsilon \mathbb{R} y \neq 0}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not for the function you have above, no.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y doesnot equal 4** sorry typo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y can equal 4 here, just not anything above it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my solution says it says y can equal any real number :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i thought it was that too,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not for a parabola it can't.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The domain is any real number for x...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that's what i thought, look at number 8 http://mail.wecdsb.on.ca/~janet_maitland/FOV10006CFBC/FOV10006E9C7/unit%20three%20practice%20answers.pdf?Plugin=Loft

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe she made a mistake :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, you can see clearly from the graph that the range is y≤4.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay thanks :D i thought i might have been doing to wrong
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