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anonymous

  • one year ago

Pre Calc Help

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  1. anonymous
    • one year ago
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  2. KaiserTheSlayer
    • one year ago
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    do want to know how or the answer??

  3. anonymous
    • one year ago
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    both please

  4. KaiserTheSlayer
    • one year ago
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    ok

  5. anonymous
    • one year ago
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    i know it is steepness but idk the exact rule

  6. anonymous
    • one year ago
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    can anyone help?

  7. KaiserTheSlayer
    • one year ago
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    yes but it would take a little

  8. anonymous
    • one year ago
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    thats fine

  9. anonymous
    • one year ago
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    I just need someone to explain how i know A based off looking at graph i know its negative but how do i know if its 1,2, etc

  10. KaiserTheSlayer
    • one year ago
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    alright, so f(x)=-(x-1)^2+4, does it make sense why

  11. anonymous
    • one year ago
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    can you give it in standard form

  12. anonymous
    • one year ago
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    ax^2 + bx + c

  13. anonymous
    • one year ago
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    and no im not sure how you got that

  14. KaiserTheSlayer
    • one year ago
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    okay so since we know the vertext, we know how much to shift it, up four and to the right 1, so thats how i got that

  15. KaiserTheSlayer
    • one year ago
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    to get the form you want, factor and simplfy it

  16. anonymous
    • one year ago
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    wait so if that was a steeper parabola what would you change

  17. KaiserTheSlayer
    • one year ago
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    the a and for the answer it is -1

  18. anonymous
    • one year ago
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    where is a in the formula you gave me?

  19. Nnesha
    • one year ago
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    alright you can find vertex point by looking at the graph right then substitute into this equation \[\huge\rm y = a(x-h)^2 +k\] vertex form of quadratic equation after that pick x and y values plug in solve for a :-)

  20. KaiserTheSlayer
    • one year ago
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    -x^2+2x+3

  21. Nnesha
    • one year ago
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    (x,y) (0,3)

  22. anonymous
    • one year ago
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    when y=0,x=-1 and 2

  23. Nnesha
    • one year ago
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    and by looking at the shape of graph u can easily find if a is negative or positive

  24. anonymous
    • one year ago
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    correction no,x=-1 and x=3

  25. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @MathMan77 where is a in the formula you gave me? \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @KaiserTheSlayer alright, so f(x)=-(x-1)^2+4, does it make sense why \(\color{blue}{\text{End of Quote}}\) \[y=-(x-1)^2+4\] --- >\[\huge\rm y = -\color{red}{a}(x-h)^2 +k\] invisble one

  26. anonymous
    • one year ago
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    ah thank you!

  27. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha alright you can find vertex point by looking at the graph right then substitute into this equation \[\huge\rm y = a(x-h)^2 +k\] vertex form of quadratic equation after that pick x and y values plug in solve for a :-) \(\color{blue}{\text{End of Quote}}\) vertex point ( 1,4) (x,y) (0,3) \[\large\rm 3=a(0-1)^2 +4\] like this if u solve for a u will get the answer

  28. anonymous
    • one year ago
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    y=-(x+1)(x-3)=-(x^2-2x-3)=-(x^2-2x)+3=-(x^2-2x+1-1)+3=-(x^2-2x+1)+1+3 =-(x-1)^2+4

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