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anonymous
 4 years ago
consider the following quadratic function...k(x)=(x6)..find the vertex of this function
anonymous
 4 years ago
consider the following quadratic function...k(x)=(x6)..find the vertex of this function

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's not a quadratic function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is (x6) raised to the 2nd power?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no...it's not on the problem...that's how it's worded

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have the answer..I just don't understand how to do the problem...the answer is (6,0)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I'm going to assume it is k(x)=(x6)^2 Expand: x^212x+36 The vertex of a parabola written in the for ax^2+bx+c is: \[(\frac{b}{2a},f(\frac{b}{2a})\] \[(\frac{12}{2},f(\frac{12}{2}))=(6,f(6))=(6,0)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0looks good....worked it out..now quick question...why is f(6)=0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0f(6)=6^212(6)+36=3672+36=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or in the original form, f(x)=(x6)^2 f(6)=(66)^2=0^2=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0makes sense now thank you
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