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anonymous

  • 5 years ago

give the vertex and x-intercepts of f(x) -x squared +3x +2

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  1. anonymous
    • 5 years ago
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    is this f(x) = x^2 +3x + 2?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    ok so to find the x intercepts we have to factor the quadratic. so if we factor the quadratic we will get the following: (x+2)(x+1) therefore x = -1, -2. so they are your x intercepts. now to find the vertex the function will be at minimum. so we will have ... to find the min or max of a function you have to take the derivative. so derivative of this quad is 2x + 3 = 0 so 2x = -3 and x = -3/2 Hence the vertex of this is (-3/2, -2) i believe.

  4. myininaya
    • 5 years ago
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    or you can write the parabola in the form f(x)=a(x-h)^2+k

  5. myininaya
    • 5 years ago
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    vertex=(h,k)

  6. myininaya
    • 5 years ago
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    You need to know how to complete the square though

  7. myininaya
    • 5 years ago
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    so f(x)=x^2+3x+(3/2)^2+2+(3/2)^2=(x+3/2)^2+2+9/4=(x+3/2)^2+8/4+9/4=(x+3/2)^2+17/4 so the vertex is (-3/2,17/4)

  8. myininaya
    • 5 years ago
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    oops but i did something wrong

  9. anonymous
    • 5 years ago
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    I was about to say myin, but you are more intelligent than I. Fix it nao! :)

  10. myininaya
    • 5 years ago
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    f(x)=x^2+3x+(3/2)^2+2-(3/2)^2=(x+3/2)^2+2-9/4=(x+3/2)^2+8/4-9/4=(x+3/2)^2-1/4

  11. myininaya
    • 5 years ago
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    so the vertex is (-3/2,1/4) if I didn't screw up again

  12. myininaya
    • 5 years ago
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    darn it (-3/2,-1/4)

  13. myininaya
    • 5 years ago
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    and laxad found the x-intercepts above

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