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anonymous

  • 4 years ago

consider the following quadratic function...k(x)=(x-6)..find the vertex of this function

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  1. anonymous
    • 4 years ago
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    that's not a quadratic function

  2. anonymous
    • 4 years ago
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    is (x-6) raised to the 2nd power?

  3. anonymous
    • 4 years ago
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    no...it's not on the problem...that's how it's worded

  4. anonymous
    • 4 years ago
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    k(x)=x-6 is a line

  5. anonymous
    • 4 years ago
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    I have the answer..I just don't understand how to do the problem...the answer is (6,0)

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

  7. anonymous
    • 4 years ago
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    looks good....worked it out..now quick question...why is f(6)=0?

  8. anonymous
    • 4 years ago
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    f(6)=6^2-12(6)+36=36-72+36=0

  9. anonymous
    • 4 years ago
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    or in the original form, f(x)=(x-6)^2 f(6)=(6-6)^2=0^2=0

  10. anonymous
    • 4 years ago
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    makes sense now thank you

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spraguer (Moderator)
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