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anonymous

  • one year ago

Which function has a vertex at the origin? f(x) = (x + 4)2 f(x) = x(x – 4) f(x) = (x – 4)(x + 4) f(x) = –x2

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  1. SolomonZelman
    • one year ago
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    For any parent function \(\large\color{black}{ \displaystyle y=f(x) }\) (assuming it goes through the origin): -------------------------------------------- \(\large\color{black}{ \displaystyle y=f(x+c) }\) is a shift \(c\) units to the left. -------------------------------------------- \(\large\color{black}{ \displaystyle y=f(x-c) }\) is a shift \(c\) units to the right. -------------------------------------------- \(\large\color{black}{ \displaystyle y=f(x)+c }\) is a shift \(c\) units up. -------------------------------------------- \(\large\color{black}{ \displaystyle y=f(x)-c }\) is a shift \(c\) units down. -------------------------------------------- \(\large\color{black}{ \displaystyle y=c{\tiny}f(x) }\) is stretching (widening) the function when \(0<c<1\) -------------------------------------------- \(\large\color{black}{ \displaystyle y=c{\tiny}f(x) }\) is making the function thiner and taller when \(c>1\) (this all is for positive number c)

  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y=c{\tiny}f(x) }\) is called multiplying the function times the scale factor, and if the parent function y=f(x) is already going through the origin, then so would be the y=c•f(x) (regardless of value of c)

  3. SolomonZelman
    • one year ago
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    One note to make: When c is a scale factor it can't equal zero, because the entire function then is going to be just a line y=0

  4. SolomonZelman
    • one year ago
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    if you got question(s), then please ask.... :)

  5. anonymous
    • one year ago
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    no questions , btw thanks

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