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Destinyyyy

  • one year ago

Find the coordinates of the vertex and the intercepts of the following quadratic function. When necessary, approximate the x-intercepts to the nearest tenth.

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  1. Destinyyyy
    • one year ago
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    f(x)+x^2+6x+5

  2. Destinyyyy
    • one year ago
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    a=1 b=6 c=5 =-(6)/2(1)

  3. princeharryyy
    • one year ago
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    y = (x+1)(x+5), I guess you can solve the rest by yourself

  4. Destinyyyy
    • one year ago
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    ??

  5. princeharryyy
    • one year ago
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    f(x)+x^2+6x+5 can be expressed as y = (x+1)(x+5)

  6. Destinyyyy
    • one year ago
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    Um okay.. This is my first problem with this. And I basically have no clue as to what im doing. And what your saying doesnt follow my examples and you haven't said anything about what ive already solved.

  7. Destinyyyy
    • one year ago
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    It says to solve for the vertex with x=-b/2a Then plug the answer for x into the problem to get the y-intercept. Factor, then set each factor to 0, that gives the answer x-intercept.

  8. princeharryyy
    • one year ago
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    it's a point where you get the maxima or minima

  9. princeharryyy
    • one year ago
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    |dw:1436900448579:dw|

  10. princeharryyy
    • one year ago
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    hope it helps!

  11. Destinyyyy
    • one year ago
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    I guess.. Now what?

  12. princeharryyy
    • one year ago
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    intercepts (-5,0) and (-1,0) for vertex put x = -3 in the equation find y y = (-3+1)(-3+5) = -4 so vertex (-3,-4)

  13. Destinyyyy
    • one year ago
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    Um why are you doing it with y? My examples show to do it like this-- f(-3)= (-3)^2 -6(-3)+5

  14. princeharryyy
    • one year ago
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    y = f(x) they are both the same

  15. princeharryyy
    • one year ago
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    we write it like this >> y = f(x) = (x+1)(x+5) y is a function in x

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